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Mathematics Chapter 5 Arithmetic Progressions

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    NCERT Solution For Class 10 Mathematics

    Arithmetic Progressions Here is the CBSE Mathematics Chapter 5 for Class 10 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 10 Mathematics Arithmetic Progressions Chapter 5 NCERT Solutions for Class 10 Mathematics Arithmetic Progressions Chapter 5 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 10 Mathematics.

    Question 1
    CBSEENMA10007384
    Wired Faculty App

    In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

    The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
    Solution

    Taxi fare for 1 km = Rs. 15 = a1 Taxi fare for 2 kms
    = Rs. 15 + Rs. 8 = Rs. 23 = a2 Taxi fare for 3 kms
    = 23 + Rs. 8 = Rs. 31 = a3 Taxi fare for 4 kms
    = Rs. 31 + Rs. 8 = Rs. 39 = a4
    and so on.
    a2 – a1 = Rs. 23 – Rs. 15 = Rs. 8
    a3 – a2 = Rs. 31 – Rs. 23 = Rs. 8
    a4 – a3 = Rs. 39 – Rs. 31 = Rs. 8
    i.e., ak + 1 – ak is the same every time.

    Question 2
    CBSEENMA10007386
    Wired Faculty App

    In which of the following situational, does the list of numbers involved make an arithmetic progression, and why?

    The cost of digging a well for the first metre is Rs. 150 and rises by Rs. 50 for each succeeding metre.

    Solution

    Cost of digging the well after 1 metre of digging of Rs. 150 = a4
    Cost of digging the well after 2 metres of digging
    = Rs. 150 + Rs. 50
    = Rs. 200 = a2
    Cost of digging the well after 3 metres of digging
    = Rs. 150 + Rs. 50
    = Rs. 2a = a3
    Cost of digging the well after 4 metres of digging
    = Rs. 200 + Rs. 50
    = Rs. 250 = a4
    and so on.
    a2 – a4 = Rs. 200 – Rs. 150 = Rs. 50
    a3 – a2 = Rs. 250 – Rs. 200 = Rs. 50
    a4 – a3 = Rs. 350 – Rs. 250 = Rs. 50
    i.e., ak +1 – ak is the same every time.
    So this list of numbers forms an A .P. with the first term a Rs. 150 and the common difference d = Rs. 50.

    Question 3
    CBSEENMA10007395
    Wired Faculty App

    In which of the following situational, does the list of numbers involved make an arithmetic progression, and why?

    The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.

    Solution

    Taxi fare for 1 km = Rs. 15 = a1 Taxi fare for 2 kms
    = Rs. 15 + Rs. 8 = Rs. 23 = a2 Taxi fare for 3 kms
    = 23 + Rs. 8 = Rs. 31 = a3 Taxi fare for 4 kms
    = Rs. 31 + Rs. 8 = Rs. 39 = a4
    and so on.
    a2 – a1 = Rs. 23 – Rs. 15 = Rs. 8
    a3 – a2 = Rs. 31 – Rs. 23 = Rs. 8
    a4 – a3 = Rs. 39 – Rs. 31 = Rs. 8
    i.e., ak + 1 – ak is the same every time.
    So, this list of numbers form an arithmetic progression with the first term a = Rs. 15 and the common difference d = Rs. 8.

    Question 4
    CBSEENMA10007404
    Wired Faculty App

    In which of the following situational, does the list of numbers involved make an arithmetic progression, and why?

    The amount of air present in a cylinder when a vacuum pump removes 1 fourth of the air remaining in the cylinder at a time.

    Solution

    Amount of air present in the cylinder = x units (say) = a1
    Amount of air present in the cylinder after one time removal of air by the vacuum pump
    equals straight x plus straight x over 4 equals fraction numerator 3 straight x over denominator 4 end fraction space units space equals space straight a subscript 2
    Amount of air present in the cylinder after two times removal of air by the vacuum pump
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    and so on,
    straight a subscript 1 minus straight a subscript 1 equals fraction numerator 3 straight x over denominator 4 end fraction minus straight x equals straight x over 4 space units
    straight a subscript 3 minus straight a subscript 2 equals open parentheses 3 over 4 close parentheses squared straight x minus 3 over 4 straight x equals fraction numerator 3 straight x over denominator 4 end fraction space units.
    As a3 – a7 ≠ a3 – a2, this list of numbers does not form an A .P.

    Question 5
    CBSEENMA10007407
    Wired Faculty App

    In which of the following situational, does the list of numbers involved make an arithmetic progression, and why?

    The cost of digging a well for the first metre is Rs. 150 and rises by Rs. 50 for each succeeding metre.

    Solution

    Cost of digging the well after 1 metre of digging of Rs. 150 = a4
    Cost of digging the well after 2 metres of digging
    = Rs. 150 + Rs. 50
    = Rs. 200 = a2
    Cost of digging the well after 3 metres of digging
    = Rs. 150 + Rs. 50
    = Rs. 2a = a3
    Cost of digging the well after 4 metres of digging
    = Rs. 200 + Rs. 50
    = Rs. 250 = a4
    and so on.
    a2 – a4 = Rs. 200 – Rs. 150 = Rs. 50
    a3 – a2 = Rs. 250 – Rs. 200 = Rs. 50
    a4 – a3 = Rs. 350 – Rs. 250 = Rs. 50
    i.e., ak +1 – ak is the same every time.
    So this list of numbers forms an A .P. with the first term a Rs. 150 and the common difference d = Rs. 50.

    Question 6
    CBSEENMA10007408
    Wired Faculty App

    In which of the following situational, does the list of numbers involved make an arithmetic progression, and why?

    The amount of money in the account every year, when ` 10000 is deposited at compound interest at 8 % per annum.

    Solution
    Amount of money after 1 year
    = Rs. 1000 <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Amount of money after2 years
    = Rs. 10000 open parentheses 1 plus 8 over 100 close parentheses squared equals straight a subscript 2
    Amount of money after 3 years
    = Rs. 10000 open parentheses 1 plus 8 over 100 close parentheses cubed equals straight a subscript 3
    Amount of money after 4 years
    = Rs. 1000 <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    a2 - a1 = Rs. 10000  open parentheses 1 plus 8 over 100 close parentheses squared minus space Rs. space 1000 space open parentheses 1 plus 8 over 100 close parentheses
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    As a2 – a4 ≠ a1– a2, this list of numbers does not form in A .P.
    Question 7
    CBSEENMA10007409
    Wired Faculty App

    Write first four terms of the AP, when the first term a and the common difference d are given as follows:
    a – 10, d = 10
    Solution

    a = 10, d = 10
    First term = a = 10
    Second term = 18 + d = 10+ 10 = 20
    Third term = 20 + d = 20 + 10 = 30
    Fourth term = 30 + d = 30 + 10 = 40
    Hence, first four terms of the given A .P. are 10, 20, 30, 40.

    Question 8
    CBSEENMA10007410
    Wired Faculty App

    Write first four terms of the AP, when the first term a and the common difference d are given as follows:
    a = –2, d = 0
    Solution

    a = –2, d = 0
    First term = a = – 2
    Second term = –2 + d = –2 + 0 = –2
    Third term = –2 + d = –2 + 0 = –2
    Fourth term = –2 + d = –2 + 0 = –2
    Hence first, first four terms of the given A .P. are -2, -3, -2 -2.

    Question 9
    CBSEENMA10007411
    Wired Faculty App

    Write first four terms of the AP, when the first term a and the common difference d are given as follows:
    a = 4, d = – 3
    Solution

    a = 4, d = –3
    First term = a = 4
    Second term = 4 + d = 4 + (–3) = 1
    Third term = 1 + d = 1 + (–3) = –2
    Fourth term = –2 + d = –2 + (–3) = –5
    Hence, four first terms of the given A .P. are 4, 1, -2, -5.

    Question 10
    CBSEENMA10007415
    Wired Faculty App

    Write first four terms of the AP, when the first term a and the common difference d are given as follows:

    a = – 1, d = 1 half
    Solution

    a = – 1, d = 1 half
    First term = a = -1
    Second term  space space space space space equals space 1 plus d equals negative 1 plus 1 half equals negative 1 half
    Third term  equals negative 1 half plus straight d equals negative 1 half plus 1 half equals 0
    Fourth termequals 0 plus straight d equals 0 plus 1 half equals 1 half
    Hence, first four terms of the given A .P. are
    negative 1 comma space minus 1 half comma space 0 comma space 1 half

    Question 11
    CBSEENMA10007416
    Wired Faculty App

    Write first four terms of the AP, when the first term a and the common difference d are given as follows:

    a = – 1.25, d = – 0.25
    Solution

    a = –1.25, d = –0.25
    First term = a = –1.25
    Second term = –1.25 + d
    = – 1.25 + (–0.25)
    = –1.50
    Third term = – 1.50 + d
    = – 1.50 + (–0.25)
    = – 1.75
    Fourth term = – 1.75 + d
    = – 1.75 + (–0.25)
    = –2.00

    Question 12
    CBSEENMA10007420
    Wired Faculty App

    For the following APs, write the first term and the common difference
    3, 1, – 1, – 3, . . .
    Solution

    3, 1, –1, –3, .... First term (a) = 3
    Common difference (d) = 1 – 3 = –2

    Question 13
    CBSEENMA10007421
    Wired Faculty App

    For the following APs, write the first term and the common difference
    – 5, – 1, 3, 7, . . .
    Solution

    –5, –1, 3, 7,.....
    First term (a) = –5
    Common difference (d) = –1 – (–5) = –1 + 5 = 4

    Question 14
    CBSEENMA10007422
    Wired Faculty App

    For the following APs, write the first term and the common difference

    1 third comma space 5 over 3 comma space 9 over 3 comma space 13 over 3 comma space space.......

    Solution

    First   term (a) = 1 third
    Common difference (d) = 5 over 3 minus 1 third equals 4 over 3

    Question 15
    CBSEENMA10007424
    Wired Faculty App

    For the following APs, write the first term and the common difference
    0.6, 1.7, 2.8, 3.9, . . .
    Solution

    0.6, 1.7, 2.8, 3.9, . . .
    First term (a) = 0.6
    Common difference =(d) = 1.7 - 0.6 = 1.1

    Question 16
    CBSEENMA10007427
    Wired Faculty App

    Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
    2, 4, 8, 16, . . .
    Solution

    2, 4, 8, 16, . . .
    a2 - a1 = 4 - 2 = 2
    a3 - a2 = 8 - 4 = 4 
    a4 - a3  = 16 - 8 = 8
    Here,   a2 - a1 space space not equal to a3 - a2 
    So the given list of numbers does not from an A .P.

    Question 17
    CBSEENMA10007430
    Wired Faculty App

    Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

    2 comma space 5 over 2 comma space 3 comma space 7 over 2 comma....
    Solution
    2 comma space 5 over 2 comma space 3 comma space 7 over 2 comma....
    a2 - a1 = 5 over 2 minus 2 equals 1 half
    a3 - a2 = 3 minus 5 over 2 equals 1 half
    a4 - a3 = 7 over 2 minus 3 equals 1 half
    i.e., ak+1 – ak is the same every time so, the given list of numbers forms an A .P. with the common
    difference d = 1 half.
    The next three terms are :
    7 over 2 plus 1 half equals 4 4 plus 1 half equals 9 over 2 and space 9 over 2 plus 1 half equals 5

    Sponsor Area

    Question 18
    CBSEENMA10007432
    Wired Faculty App

    Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
    – 1.2, – 3.2, – 5.2, – 7.2, . . .
    Solution

    –1.2, –3.2, –5.2, –7.2
    a2 – a4 = –3.2 – (–1.2)
    = –3.2 + 1.2
    = –2.0
    a3 – a2 = –5.2 – (–3.2)
    = –5.2 + 3.2
    = –2.0
    a3 – a4
    = –7.2 – (–5.2)
    = –7.2 + 5.2
    = –2.0
    i.e., ak+1 – ak is the same every time. So, the given list of numbers forms an A .P. with the common difference d = – 2.0
    The next three terms are :
    – 7.2 + (–2.0) = –9.2
    –9.2 + (–2.0) = –11.2
    and –11.2 + (–2.0) = –13.2

    Question 19
    CBSEENMA10007433
    Wired Faculty App

    Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
     –10, –6, –2, 2, ....
    Solution

     –10, –6, –2, 2, ....
    a2 – a1 = – 6– (–10)
    = –6 + 10 = 4
    a3 – a2 = –2 – (–6)
    = –2 + 6 = 4
    a4 – a3 = 2 – (–2)
    = 2 + 2 = 4
    i.e., ak + 1 – ak is he same every time. So, the given list of numbers forms an A .P. with
    the common difference d = 4.
    The next three terms are :
    2 + 4 = 6, 6 + 4= 10
    and 10 + 4 = 14

    Question 20
    CBSEENMA10007439
    Wired Faculty App

    Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Solution
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
     a subscript 2 minus a subscript 1 space equals open parentheses 3 plus square root of 2 close parentheses minus 3 equals square root of 2 straight a subscript 3 minus straight a subscript 2 equals open parentheses 3 plus 2 square root of 2 close parentheses minus left parenthesis 3 plus square root of 2 right parenthesis space space space space space space space space space equals square root of 2 straight a subscript 4 minus straight a subscript 3 equals open parentheses 3 plus 3 square root of 2 close parentheses minus left parenthesis 3 plus 2 square root of 2 right parenthesis space space space space space space space space space space space equals space square root of 2
    i.e., ak + 1 – ak is the same every time. So, the eiven list of numbers forms an A .P. with the common difference  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    The next three terms are :
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Question 21
    CBSEENMA10007442
    Wired Faculty App

    Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
    0.2, 0.22, 0.222, 0.2222, . . .
    Solution

    0.2, 0.22, 0.222, 0.2222, ......
    a2 – a1 = 0.22 – 0.2 = 0.02
    a3 – a2 = 0.222 – 0.22 = 0.002
    As a2 – a4 ≠ a3 – a2
    the given list of numbers does not form an A.P.

    Question 22
    CBSEENMA10007444
    Wired Faculty App

    Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
    0, – 4, – 8, –12, . . .
    Solution

    0, –4, –8, –12,......
    a2 – a1 = –4 – 0 = –4
    a3 – a2 = – 8 – (–4)
    = – 8 + 4 = –4
    a4 – a3 = –12 – (–8) = –12 + 8 – 5
    i.e., ak + 1 – ak is the same every time. So. the given list of numbers forms an A .P. with the common difference d = –4.
    The next three terms are :
    – 12 + (–4) = –12 – 4 = –16
    – 16 + (–4) = –16 – 4 = –20
    and – 20 + (–4) = –20 – 4 = –24

    Question 23
    CBSEENMA10007449
    Wired Faculty App

    Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

    negative 1 half comma space minus 1 half space. space minus 1 half space. negative 1 half space comma.....
    Solution
    negative 1 half comma space minus 1 half space. space minus 1 half space. negative 1 half space comma..... straight a subscript 2 minus straight a subscript 1 equals negative 1 half comma space minus open parentheses negative 1 half close parentheses space equals negative 1 half plus 1 half equals 0 straight a subscript 3 minus straight a subscript 2 equals negative 1 half minus open parentheses negative 1 half close parentheses equals negative 1 half plus 1 half equals 0 straight a subscript 4 minus straight a subscript 3 equals negative 1 half space minus open parentheses negative 1 half close parentheses equals negative 1 half plus 1 half equals 0

    i.e., ak + 1 – ak is the same every time. So, the given list of numbers forms an A .P. with the common difference d = 0.
    The next three terms are :

    minus 1 half plus 0 equals negative 1 half comma space plus negative 1 half space plus space 0 space equals space minus 1 half and space space space minus 1 half space plus 0 equals space minus 1 half
    Question 24
    CBSEENMA10007450
    Wired Faculty App

    Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
    1, 3, 9, 27, . .
    Solution

    1, 3, 9, 27,.......
    a2 – a1 = 3 – 1 = 2
    a3 – a2 = 9 – 3 = 6
    As a2 – ≠ a3 – a2
    the given list of numbers does not form an A .P.

    Question 25
    CBSEENMA10007451
    Wired Faculty App

    Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
    a, 2a, 3a, 4a, . .
    Solution

    a, 2a, 3a, 4a, ........
    a2 – a1 = 2a – a = a
    a3 – a2 = 3a – 2a = a
    a4 – a3 = 4a – 3a = a
    i.e., ak + 1 – ak is the same every time. So, the given list of numbers form an A .P. with the common difference d = 0.
    The next three terms are :
    4a + a = 5a, 5a + a = 6a
    and 6a + a = 7a.

    Question 26
    CBSEENMA10007452
    Wired Faculty App

    Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
    a, a2 , a3 , a4 , . . .
    Solution

    a, a2, a3, a4, ..............
    a2 – a1 = a2 – a = a(a – 1)
    a3 – a2 = a3 – a2 = a2 (a – 1)
    As a2 – a1 ≠ a3 – a2
    the given list of numbers does not form an A .P.

    Question 27
    CBSEENMA10007456
    Wired Faculty App

    Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Solution
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    straight a subscript 2 minus straight a subscript 1 equals square root of 8 minus square root of 2 equals 2 square root of 2 minus square root of 2 equals square root of 2 straight a subscript 3 minus straight a subscript 2 equals square root of 18 minus square root of 8 equals 3 square root of 2 minus 2 square root of 2 equals square root of 2 straight a subscript 4 minus straight a subscript 3 equals square root of 32 minus square root of 18 equals 4 square root of 2 minus 3 square root of 2 equals square root of 2
    i.e., ak + 1 ak is the same every time. So, the given list of numbers forms an A .P. with the common
    difference d = square root of 2
    The next three terms are :
    Question 28
    CBSEENMA10007461
    Wired Faculty App

    Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

    square root of 3 comma space square root of 6 comma space square root of 9 comma space square root of 12 comma space.....
    Solution
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    As a2 – a1 ≠ a3 – a2, the given list of numbers does not form an A .P.
    Question 29
    CBSEENMA10007462
    Wired Faculty App

    Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
    12, 32, 52, 72,.......
    Solution
    12, 32, 52, 72,.......
    a2 – a1 = 32 – 12 = 9 – 1 = 8
    a3 – a2 = 52 – 32 = 25 – 9 = 16
    As a2 – a1 ≠ a3 – a2, the given list of numbers does not form an A .P.
    Question 30
    CBSEENMA10007463
    Wired Faculty App

    Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
    12, 52, 72, 73,....... 
    Solution

    12, 52, 72, 73,.......
    = 52 – 12 = (5 – 1) (5 + 1)
    = (4) (6) = 24
    a3 – a2 = 72 – 52 = (7 – 5) (7 + 5)
    = (2) (12) = 24
    a4 – a3 = 73 – 72 = 73 – 49 = 24
    i.e., ak + 1 ak is the same every time. So, the given list of numbers forms an A .P. with
    the common difference d = 24.
    The next three terms are : 73 + 24 = 97 97 + 24 = 121
    and 121 + 24 = 145.

    Question 31
    CBSEENMA10007476
    Wired Faculty App

    Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:
    Solution
      (i)                      an = a + (n - 1)d
    Question 32
    CBSEENMA10007484
    Wired Faculty App

    Choose the correct choice in the following and justify :
    30th term of the AP: 10, 7, 4, . . . , is
    (A)   97    (B)    77     (C)   –77    (D)   – 87
    Solution

    The given A. P. is   10, 7, 4
    Here,    a = 10
                d = 7 - 10 = -3
    and       n = 30
    We know that,  an = 1 + (n - 1) d
    So,   a30 = 10 + (30 - 1) (-3)
    rightwards double arrow   a30 = 10 - 87
    rightwards double arrow   a30 = -77
    Hence the correct choice is (C) = - 77.
                 

    Question 33
    CBSEENMA10007491
    Wired Faculty App

    Choose the correct choice in the following and justify :
    30th term of the AP: 10, 7, 4, . . . , is

    11th term of the A.P. is space space minus 3 comma space 1 half comma space 2 comma space....
    Solution

    The given A. P. is space space minus 3 comma space 1 half comma space 2 comma space....
    Here,    a = -3
     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    and        n =  11
    We know that an = a + (n - 1) d
    S o comma space space space space space space space space space a subscript 11 equals negative 3 plus open parentheses 11 minus 1 close parentheses open parentheses 5 over 2 close parentheses rightwards double arrow space space space space space space space space space space a subscript 11 equals negative 3 plus 25 rightwards double arrow space space space space space space space space space space a subscript 11 space equals space 22
    Hence, the correct choice is (B) 22. 

    Question 34
    CBSEENMA10007505
    Wired Faculty App

     In the following APs, find the missing terms in the boxes.

    2. space square. space 26 space
    Solution

     Let the common difference of the given A .P. be d. Then,
    Third term = 2 + d + d= 2 + 2d According to the question,
             2 + 2d = 26
    rightwards double arrow space space space space space2d = 26 - 2
    rightwards double arrow    2d = 24
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    So, second term = 2 + d = 2 + 12 = 14.
    Hence, the missing termed in the box 14.



    Question 35
    CBSEENMA10007509
    Wired Faculty App

    In the following APs, find the missing terms in the boxes.

    square comma space 13. space square. space 3



    Solution

    Let the first term and the common difference of the given A .P. be a and b respectively.
    Second term = 13
    ⇒ a + (2 – 1) d = 13
    ⇒ a + d = 13 ...(i)
    Fourth term = 3
    ⇒ a+ (4 – 1)d = 3
    ⇒ a + 3d = 3 ...(ii)
    Solving (i) and (ii), we get
    a = 18
    d = –5
    Therefore,
    Third term = a + (3 – 1)d
    = a + 2d
    = 18 + 2(–5)
    = 18 –10 = 8
    Hence, the missing terms in the boxes are space space space space box enclose 18 space and space box enclose 8.

    Question 36
    CBSEENMA10007515
    Wired Faculty App

    In the following APs, find the missing terms in the boxes.

    5 comma space square comma space square comma space 9 1 half
    Solution
    Let the common difference of the given A .P. be d.
                       a = 5
                4th term = 9 1 half
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>  5 + (4 - 1) d = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    rightwards double arrow space space space space space space space 3 straight d space equals space 19 over 2 minus 5 rightwards double arrow space space space space space space space 3 straight d space equals space 9 over 2 rightwards double arrow space space space space space space space space straight d equals space 3 over 2
    Therefore,
     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Hence, missing terms in the boxes are <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Question 37
    CBSEENMA10007520
    Wired Faculty App

    In the following APs, find the missing terms in the boxes.

    negative 4 comma space square comma square comma space square comma square comma space 6
    Solution

    Let the common difference of the given A. P. be d.
                             a = - 4
                      6th term = 6
    rightwards double arrow space space minus 4 plus left parenthesis 6 minus 1 right parenthesis straight d space equals 6 space space space space space space space space space space space space space left square bracket because space straight a subscript straight n equals straight a plus open vertical bar straight n minus 1 close vertical bar straight d right square bracket rightwards double arrow space space minus 4 plus 5 straight d equals 6 rightwards double arrow space space space space space space 5 straight d equals 6 plus 4 rightwards double arrow space space space space space space space 5 straight d space equals space 10 rightwards double arrow space space space space space space space space straight d equals 10 over 5 rightwards double arrow space space space space space space space space straight d space equals space 2
    Therefore,
          Second term = -4 + 2 = -2
          Third term = -2 + 2 = 0
          Fourth term = 0 + 2 = 2
    and, Fifth term = 2 + 2 = 4
    Hence, the missing terms in the boxes are
    box enclose negative 2 end enclose comma space box enclose 0 comma space box enclose 2 comma space box enclose 4

    Question 38
    CBSEENMA10007523
    Wired Faculty App

    In the following APs, find the missing terms in the boxes.

    square comma space 38 comma space square comma space square comma space square comma space minus 22.
    Solution

    Let the first terms and the common difference of the given A .P. be a and d respectively.
    Second term = 38
    ⇒ a + (2 – 1)d = 38
    [∴ an = a + (n – 1) d]
    ⇒ a + d = 38 ...(i)
    ⇒ Sixth term = –22
    ⇒ a + (6 – 1 )d = –22
    ⇒ a + 5d = –22 ...(ii)
    Solving (i) and (ii), we get
    a = 53 d = –15
    Therefore,
    Third term = 53 + (3 –1 (–5)
    [∵ an = a + (n – 1) d] = 53 – 30 = 23
    Fourth term = 53 + (4 – 1) (–15)
    [∵ an = a + (n – 1) d] = 8
    Fifth term = 53 + (5 – 1) (–15)
    [∵ = a + (n – 1) d] = –7
    Hence, the missing terms in the boxes are :
    box enclose 53 comma space box enclose 23 comma space box enclose 8 comma space box enclose negative 7 end enclose.

    Question 39
    CBSEENMA10007527
    Wired Faculty App

    Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
    Solution

    The given A. P. is 3, 8, 13, 18, .....
    Here,     a = 3
                 d = 8 - 3 = 5
    Let the nth term f=of the A. P. be 78
    Then,   an = a + (n - 1)d
    space space rightwards double arrow space space space  78 = 3 + (n -1) (5)
     rightwards double arrow     5(n - 1) = 78 - 3
     rightwards double arrow     5 (n - 1) = 75
     rightwards double arrow space space space space space space n minus 1 equals 75 over 5        
     rightwards double arrow     n - 1 = 15
     rightwards double arrow     n = 15 + 1
     rightwards double arrow     n = 16
    Hence, 16th term of the A .P. is 78.
     
                            

    Sponsor Area

    Question 40
    CBSEENMA10007533
    Wired Faculty App

    Find the number of terms in each of the following APs :
    7, 13, 19, . . . , 205
    Solution

    7, 13, 19, . . . , 205
    Here,      a = 7
                  d = 13 - 7 = 6
    And,        an = 205
    Let the number of terms be n.
    Then,    an = 205
    rightwards double arrow  a + ( n - 1) d = 205
    rightwards double arrow  7 + ( n - 1)6 = 205
    rightwards double arrow  6(n - 1) = 205 - 7
    rightwards double arrow  6(n - 1) = 198
    rightwards double arrow space space space space space n minus 1 equals 198 over 6
    rightwards double arrow    n - 1 = 33
    rightwards double arrow    n = 33 + 1
    rightwards double arrow    n = 34
    Hence, the number of terms of the given A .P., is 34.

    Question 41
    CBSEENMA10007534
    Wired Faculty App

    Find the number of terms in each of the following APs :

    space space space 18 comma space 15 1 half comma space 13 comma space....... comma space minus 47
    Solution
    bold space bold space bold space bold 18 bold comma bold space bold 15 bold 1 over bold 2 bold comma bold space bold 13 bold comma bold space bold. bold. bold. bold. bold. bold. bold. bold comma bold space bold minus bold 47
    Here,    a= 18
                straight d space equals space 15 1 half minus 18 equals 31 over 2 minus 18 equals negative 5 over 2
    And,      an = -47
    Let the number of terms be n.
    Then,      an = -47
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>  a + (n - 1)d = - 47rightwards double arrow space 18 plus left parenthesis straight n minus 1 right parenthesis open parentheses fraction numerator negative 5 over denominator 2 end fraction close parentheses equals negative 47 rightwards double arrow space space fraction numerator negative 5 over denominator 2 end fraction left parenthesis straight n minus 1 right parenthesis equals negative 47 minus 18 rightwards double arrow space space space space fraction numerator negative 5 over denominator 2 end fraction left parenthesis straight n minus 1 right parenthesis equals negative 65 rightwards double arrow space space space space space 5 over 2 left parenthesis straight n minus 1 right parenthesis space equals space 65 rightwards double arrow space space straight n minus 1 space equals space fraction numerator 65 cross times 2 over denominator 5 end fraction rightwards double arrow space space straight n space minus space 11 space equals space 26 rightwards double arrow space space straight n space equals space 26 space plus space 1 space rightwards double arrow space space straight n space equals space 27
    Hence, the number of terms of the given A .P. is 27.
    Question 42
    CBSEENMA10007540
    Wired Faculty App

    Check whether – 150 is a term of the AP : 11, 8, 5, 2 . .
    Solution

    The given list of number is
    11, 8, 5, 2,.....
    a2 – ax = 8 – 11 = –3
    a2 – a2 = 5 – 8 = –3
    And, a4 – a3 = 2 – 5 = –3
    i.e., ak + 1 – ak is the same every time. So, the given list of numbers form an A .P. with first term a = 11 and the common difference d = –3.
    Let –150 be the nth term of the given A .P.
    Then,         an = -150
     rightwards double arrow a + (n - 1)d = - 150
    rightwards double arrow 11 + (n - 1) (-3) = - 150
    rightwards double arrow space-3(n - 1 ) = - 150 -11
    rightwards double arrow -3 (n - 1) = -161
    rightwards double arrow 3 ( n - 1) = - 161
    rightwards double arrow 3(n - 1) = 161
    rightwards double arrow space space straight n minus 1 equals 161 over 3

    Question 43
    CBSEENMA10007543
    Wired Faculty App

    Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
    Solution
    Here,  a11 = 38 and a16 = 38
    Question 44
    CBSEENMA10007544
    Wired Faculty App

    An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term. 
    Solution
                                          a3 = 12
    Question 45
    CBSEENMA10007550
    Wired Faculty App

    If the 3rd and 9th terms of an A .P. are 4 and -8 respectively, which term is zero?

    Solution
    Here       a3  = 4

    Hence, 5th term of the given A .P. is zero.
    Question 46
    CBSEENMA10007553
    Wired Faculty App

    The 17th term of an A .P. exceeds the 10th term by 7. Find the common difference.

    Solution

    We have
    a17 = a + 16d
    and a10 = a + 9d
    According to the given question,
    a17 – a10 = 7
    ⇒ (a + 16d) – (a + 9d) = 7
    ⇒ a + 16d – a – 9d = 7
    ⇒ 7d – 7
    ⇒ d = 1
    Hence, common difference is 1.

    Question 47
    CBSEENMA10007556
    Wired Faculty App

    Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
    Solution

    Given A.P. is : 3, 15,27,39.....
    Here a1 = 3, a2 = 15
    a3 = 27, a4 = 39
    d = a2 – a1 = 15 – 3 = 12
    a54 = a (54 – 1)d
    = 3 + 53 x 12
    = 3 + 636 = 639
    Let ntn term be 132 more than a54 We know that,
    an = a + (n - 1)d
    a54 + 132 =  a + (n - 1)d
    rightwards double arrow 639 + 132 = 3 + (n - 1) x 12
    rightwards double arrow 771 - 3 = 12(n - 1)
    rightwards double arrow space 768 = 12(n - 1)
    rightwards double arrow space straight n minus 1 equals 768 over 12 equals 64 rightwards double arrow space straight n space equals space 64 space plus space 1 space equals space 65

    Question 48
    CBSEENMA10007557
    Wired Faculty App

    Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
    Solution

    Let the common difference of both the
    A.P.’s = d
    and First term of Ist A.P. = a
    also first term of 2nd A .P. = b
    Then, 200th term of 1st A .P. = a + 99d
    and 100th term of IInd A .P. = b + 99d
    Difference between their 100th term = 100
    (given)
    ⇒ (a + 99d) – (b + 99d) = 100
    ⇒ a – b = 100 ...(i)
    Now, a1000 of Ist A.P. = a + 999d
    and a1000 of Ilnd A .P. = b + 999d
    Hence, difference between their 1000th terms
    = (a + 999d) (b + 999d) = a – b
    = 100. [Using (i)]

    Question 49
    CBSEENMA10007558
    Wired Faculty App

    How many three-digit numbers are divisible by 7?

    Solution
    Three-digit numbers divisible by 7 are:
    105, 112, 119,  ...., 994.
    Here,         a = 105
                     d = 112 - 105 = 7
    and            an = 994
    We know that,
                    an = a + (n -1)d
    rightwards double arrow   994 = 105 + (n - 1) x 7
    rightwards double arrow   994 - 105 = 7 x (n - 1)
    rightwards double arrow   889 = 7 x (n - 1)
    rightwards double arrow space space space space space space space straight n minus 1 space space space equals space 889 over 7 equals 127 rightwards double arrow space space space space space space space space space space straight n space equals space 127 space plus space 1 space equals space 128
    Hence, 128 three digit numbers are divisible by 7.
    Question 50
    CBSEENMA10007560
    Wired Faculty App

    How many multiples of 4 lies between 10 and 250?

    Solution
    Multiples of 4 lying between 10 and 250 are :

    Hence, 60 multiples of 4 lies between 10 and 250.
    Question 51
    CBSEENMA10007561
    Wired Faculty App

    For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
    Solution

    Given A .P.s are :
    63, 65, 67, ... and 3, 10, 17, ...
    Here, a1 = 63, a2 = 65, a3 = 67, ....
    and 6, = 3, b2 = 10, b3 = 17.....
    Let d1 and d2 be the common differences of two given A .P.s respectively. Then
    d1 = 65 – 63 = 67 – 65 = 2
    and d2 = 10 – 3 = 17 – 10 = 7
    Now, nth term of the 1st A .P. = a1 + (n – 1)d1
    and nth term of the Ilnd A.P. = b1 + (n – 1 )d2
    ∵ nth terms of the given two A .P.s are equal.
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    rightwards double arrow 63 + (n - 1) x 2 = 3 + (n - 1) x 7
    rightwards double arrow 63 + 2n -2 = 3 + 7n - 7
    rightwards double arrow 2n + 61 = 7n - 4
    rightwards double arrow 7n - 2n = 61 + 4
    rightwards double arrow   5n = 65
    rightwards double arrow   straight n space equals space 65 over 5 equals 13
    Hence, the required value of n = 13.

    Question 52
    CBSEENMA10007562
    Wired Faculty App

    Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
    Solution

    We have
    3, 8, 13,.....253
    Here, a = 3
    d = 5 l = 253
    and n = 20
    We know that,
    nth term from the end is given by l – (n – 1)d i.e., 20th term from the end
    = 253 – (20 – 1)5
    = 253 – (19 x 5)
    = 253 – 95 = 158

    Question 53
    CBSEENMA10007564
    Wired Faculty App

    The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.

    Solution

    Case I.
    T4 + T8 = 24
    ⇒ a + (4 – 1)d + a + (8 – 1)d = 24
    ⇒ a + 3d + a + 7d = 24
    ⇒ 2a + 10d = 24
    ⇒ a + 5d = 12 ...(i)
    Case II. T6 + T10 = 44
    ⇒ a + (6 – 1)d + a + (10 – 1)d = 44
    ⇒ a + 5d + a + 9d = 44
    2a + 14d = 44
    ⇒ a + 7d = 22 ...(ii)
    Subtracting (i) from (ii), we get
    (a + 7d) – (a + 5d) = 22 – 12
    ⇒ a + 7d – a – 5d = 10
    ⇒ 2d = 10
    ⇒ d = 5

    Question 54
    CBSEENMA10007565
    Wired Faculty App

    Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000.

    Solution

    Yearly income of Subba Rao is following :
    5000, 5200, 5400, 5600...............
    Here, a = 5000, d = 200, Tn = 7000
    We know that, Tn = a + (n – 1) d
    ⇒ 7000 = 5000 + (n – 1) 200
    ⇒ 7000 = 5000 + 200n – 200
    ⇒ 7000 = 4800 + 200n
    ⇒ 200n = 2200
    ⇒ n = 11
    Hence, in the 11th year, his income reached by Rs. 7000.

    Question 55
    CBSEENMA10007566
    Wired Faculty App

    Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75, if in the nth week, her weekly savings became Rs. 20.75, find n.

    Solution

    Weekly savings of Ramkali is following :
    5, 6.75, 8.50, 10.25............
    Here, a = 5, d = 1.75, Tn = 20.75
    We know that, Tn = a + (n - 1) d
    rightwards double arrow 20.75 = 5 + (n - 1) (1.75)
    rightwards double arrow 20.72= 5 + 1.75 n - 1.75
    bold rightwards double arrow 20.75 = 1.7n + 3.25
    rightwards double arrow  1.75n = 17.5
    rightwards double arrow    straight n equals fraction numerator 17.5 over denominator 1.75 end fraction equals 10  
    Hence,    n = 10.

    Question 56
    CBSEENMA10007567
    Wired Faculty App

    Find the sum of the following APs:
    2, 7, 12, ... to 10 terms.
    Solution

    Here, a = 2
    d = 7 – 2 = 12 – 7 = 5
    And, n = 10
    We know that,
    space space space space space space straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket therefore space space straight S subscript 10 equals 10 over 2 left square bracket 2 cross times 2 plus left parenthesis 10 minus 1 right parenthesis cross times 5 right square bracket
              = 5 x [4 + 9 x 5]
              = 5 x [4 + 45]
              = 5 x 49 = 245

    Hence, the sum of First 10 terms of the given
    A .P. = 245 F

    Question 57
    CBSEENMA10007568
    Wired Faculty App

    Find the sum of the following APs:
    –37, –33, –29, . . ., to 12 terms.
    Solution

    Here, a = –31
    d = –33 – (–37)
    = – 33 + 37 = 4
    n = 12
    We know that,
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    = 6 [-74 + 11 x 4]
    = 6 x [-74 + 44]
    = 6 x (-30) = -180

    Hence, the sum of First 12 terms of the given
    A.P. = –180


    Question 58
    CBSEENMA10007569
    Wired Faculty App

    Find the sum of the following APs:
    0.6, 1.7, 2.8, . . ., to 100 terms.
    Solution

    Here a = 0.6
    d = 1.7 – 0.6 = 1.1 And, n = 100 We know that,
    straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket therefore space space straight S subscript 100 equals 100 over 2 left square bracket 2 cross times 0.6 plus left parenthesis 100 minus 1 right parenthesis cross times 1.1 right square bracket
                       
                  = 50 x [2 x 0.6 + (100 - 1) x 1.1]
                  = 50 x [1.2 + 108.9]
                  = 50 x 110.1 = 5505.0 

    Hence, the sum of first 100 terms of the given
    A.P. = 5505.


                  

    Question 59
    CBSEENMA10007572
    Wired Faculty App

    Find the sum of the following APs:

    1 over 15 comma space 1 over 12 comma space 1 over 10 comma space..... space to space 11 space terms.
    Solution

    Here,     straight a space equals space 1 over 15
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    And     n =11
    We know that,
         straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket therefore space straight S subscript 11 equals 11 over blank 2 open square brackets 2 cross times 1 over 15 plus left parenthesis 11 minus 1 right parenthesis cross times 1 over 60 close square brackets space space space space space space space space space equals 11 over 2 cross times open square brackets 2 over 15 plus 10 cross times 1 over 60 close square brackets space space space space space space space space space space equals 11 over 2 cross times open square brackets 2 over 15 plus 1 over 6 close square brackets equals 11 over 2 cross times open square brackets fraction numerator 4 plus 5 over denominator 30 end fraction close square brackets space space space space space space space space space space equals 11 over 2 cross times 9 over 30 equals 11 over 2 cross times 3 over 10 equals 33 over 20
    Hence, the sum of first 11 terms of the given  straight A. space straight P. space equals 33 over 20.

    Question 60
    CBSEENMA10007574
    Wired Faculty App

    Find the sums given below:


    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    Solution
    Let V be first term and be the common difference then,
                     a = 7
                   straight d equals 21 over 2 minus 7
                    space space space space space equals fraction numerator 21 minus 14 over denominator 2 end fraction equals 7 over 2
    We know that, an = a + (n -1)d      space space space space space space space space space space space open square brackets because space open parentheses straight a subscript straight n equals 1 close parentheses close square brackets
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    We know that,  
     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Question 61
    CBSEENMA10007576
    Wired Faculty App

    Find the sums given below:
    34 + 32 + 30 + . . . + 10

    Solution

    Let ‘a’ be the firs term and ‘d’ the common difference, then
    a = 34, d = –2, an = 10 = l
    We know that,an = a + (n – 1)
    ⇒ 10 = 34 + (n – 1) (–2)
    ⇒ 10 = 34 –2n + 2
    ⇒ 10 = 36 – 2n
    ⇒ –26 = –2n ⇒ n = 13
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    Question 62
    CBSEENMA10007577
    Wired Faculty App

    Find the sums given below:
    –5 + (–8) + (–11) + . . . + (–230)

    Solution
    Let ‘a’ be the first term and d the common difference then,
    Question 63
    CBSEENMA10007580
    Wired Faculty App

    In an AP
    Given a = 5, d = 3, an =50, find n and Sn.
    Solution

    Here a = 5, d = 3, an = 50
    We know that, an = a + (n – 1 )d
    ⇒ 50 = 5 + (n – 1) x 3
    ⇒ 50 – 5 = 3 x (n – 1)
    ⇒ 45 = 3 x (n – 1)
    rightwards double arrow space space straight n space minus space 1 space equals space 45 over 3 equals 15 rightwards double arrow space space straight n space equals space 15 space plus space 1 space equals space 16 And comma space space space space space straight S subscript straight n space equals space straight n over 2 left square bracket 2 straight a space plus space left parenthesis straight n minus 1 right parenthesis straight d right square bracket therefore space space space straight S subscript 16 equals 16 over 2 left square bracket 2 cross times 5 plus left parenthesis 16 minus 1 right parenthesis cross times 3 right square bracket space
          = 8 x [ 10 + 15 x 3]
          = 8 x 10 + 45
          = 8 x 55 = 440
    Hence,   n = 16
    and       Sn =S16 = 440

    Question 64
    CBSEENMA10007583
    Wired Faculty App

    In an AP
    given a = 7, a13 =35, find d and S13.
    Solution

    Here,  a = 7, a13 =35, find d and S13.
    We know that, an = a + (n - 1)d
    rightwards double arrow  a13 = 7 + (13 - 1)d
    rightwards double arrow  35 = 7 + 12d
    rightwards double arrow   35 - 7 = 12d
    rightwards double arrow   12d = 28
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>


    And comma space space space space straight S subscript 13 equals 13 over 2 open square brackets 2 cross times 7 plus left parenthesis 13 minus 1 right parenthesis cross times 7 over 3 close square brackets space space space space space space space space space space space space space space space space space space space space open square brackets because space straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d close square brackets space space space space space space space space space space space space space space space equals 13 over 2 left square bracket 14 plus 28 right square bracket space space space space space space space space space space space space space space space equals space 13 over 2 cross times 42 equals 13 cross times 21 equals 273 Hence comma space space space space straight d space space space equals 7 over 3 space and space straight S subscript 13 equals 273
    Question 65
    CBSEENMA10007585
    Wired Faculty App

    In an AP
    Given a12 = 37, d = 3, find ‘a’ and S12.
    Solution

    Here,    a12 = 37,   d= 3
    We know that,
          an = a + (n - 1)d
    rightwards double arrow a12 = a + (12 - 1) x 3
    rightwards double arrow 37 = a + 11 x 3
    rightwards double arrow  a = 37 - 33 = 4
    And,     straight S subscript straight n space equals straight n over 2 straight x space left parenthesis straight a plus straight l right parenthesis therefore space space space straight S subscript 12 equals 12 over 2 straight x space left parenthesis 4 plus 37 right parenthesis space space space space space space equals space 6 straight x space 41 space equals space 246 Hence comma space space space straight a space equals space 4 and space straight S subscript 12 space equals space 246 end subscript 

    Question 66
    CBSEENMA10007587
    Wired Faculty App

    Given a3 = 75, S10 = 120, find ‘d’ and a10.

    Solution

    Here,  a3 = 15
             S10  = 120
    We know that,
           straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket therefore space space straight S subscript 10 equals 10 over 2 space left square bracket 2 straight a plus left parenthesis 10 minus 1 right parenthesis straight d right square bracket rightwards double arrow space 120 space equals space 5 space straight x space left square bracket 2 straight a plus 9 straight d right square bracket rightwards double arrow 2 straight a plus 9 straight d equals 120 over 5 equals 24 rightwards double arrow space 2 straight a plus 9 straight d space equals space 24 space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis And. space space space straight a subscript 1 equals 15 rightwards double arrow space space straight a space plus 2 straight d space equals space 15 space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis
    Solving (i) and (ii)
               5d = -6
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    rightwards double arrow space space space straight a space equals space 15 plus 12 over 5 equals fraction numerator 75 plus 12 over denominator 5 end fraction equals 87 over 5 Now comma space space space space straight a subscript 10 equals straight a plus 9 straight d equals 87 over 5 plus 9 open parentheses fraction numerator negative 6 over denominator 5 end fraction close parentheses space space space space space space space space space space space space space space space space space space space space equals 87 over 5 minus 54 over 5 equals fraction numerator 87 minus 54 over denominator 5 end fraction equals 33 over 5 Hence comma space space straight d space equals space fraction numerator negative 6 over denominator 5 end fraction space and space straight a subscript 10 space equals space 33 over 5

    Question 67
    CBSEENMA10007589
    Wired Faculty App

    In an AP
    Given d = 5, 59 = 75, find ‘a’ and ‘a’
    Solution

    Here,    d = 5, S9 = 75
    We know that, 
       straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket therefore space space space space straight S subscript 9 space equals space 9 over 2 space straight x space left square bracket 2 straight a plus left parenthesis 9 minus 1 right parenthesis space straight x space 5 right square bracket rightwards double arrow space space space 75 space equals 9 over 2 straight x space left square bracket 2 straight a plus 8 space straight x space 5 right parenthesis rightwards double arrow 150 space equals space 9 cross times left square bracket 2 straight a plus 40 right square bracket rightwards double arrow 2 straight a plus 40 equals 150 over 9 equals 50 over 3 rightwards double arrow 2 straight a space equals space 50 over 3 minus 40 equals fraction numerator 50 minus 120 over denominator 3 end fraction rightwards double arrow space straight a equals fraction numerator negative 70 over denominator 3 cross times 2 end fraction equals negative 35 over 3 We space know space that. space straight a subscript straight n equals straight a plus left parenthesis straight n minus 1 right parenthesis straight d therefore space space space straight a subscript 9 equals negative 35 over 3 plus left parenthesis 9 minus 1 right parenthesis straight x 5 space space space space space space space space space equals negative 35 over 3 plus left parenthesis 9 minus 1 right parenthesis cross times 5 space space space space space space space space space space equals fraction numerator negative 35 over denominator 3 end fraction plus 40 space space space space space space space space space space space space space equals fraction numerator negative 35 plus 120 over denominator 3 end fraction equals 85 over 3 Hence comma space space space straight a equals fraction numerator negative 35 over denominator 3 end fraction space and space straight a subscript 9 equals 85 over 3

    Question 68
    CBSEENMA10007591
    Wired Faculty App

    Find the sum of the series 5 + 7 + 9+ 10 + 13 + 13 + 17 + 16 +.........to 40 terms.

    Solution
    Solution not provided.
    Ans. 1570
    Question 69
    CBSEENMA10007592
    Wired Faculty App

    Find four numbers in A .P. whose sum is 20 and the sum of whose squares is 180.

    Solution
    Solution not provided.
    Ans. –1,3,7, 11
    Question 72
    CBSEENMA10007596
    Wired Faculty App

    Three numbers are in A .P. if the sum of these numbers is 27 and the product 585, find the numbers.

    Solution
    Solution not provided.
    Ans. 9, 13, 17,21 or 9, 5, 1,-3

    Tips: -

    [Hints : Numbers are a – d, a, a + d.]
    Question 73
    CBSEENMA10007597
    Wired Faculty App

    The sum of three consecutive terms in an A .P. is 18 and their product is 192. Find the numbers.

    Solution
    Solution not provided.
    Ans. 4, 6, 8 or 8, 6, 4 16. 1,4,7 
    Question 74
    CBSEENMA10007598
    Wired Faculty App

    Which term of the A .P. : 121, 117, 113..........is its first negative term.

    Solution
    Solution not provided.
    Ans. 32rd term
    Question 76
    CBSEENMA10007600
    Wired Faculty App

    The angles of a triangle are in A .P. if the greatest angle is double the least. Find all the angles.

    Solution
    Solution not provided.
    Ans. 40°, 60°, 80° 
    Question 79
    CBSEENMA10007648
    Wired Faculty App

    Find the A .P. whose nth term is an = 10– 3n.

    Solution

    Solution not provided.
    Ans. 7, 4, 1

    Sponsor Area

    Question 80
    CBSEENMA10007649
    Wired Faculty App

    Find V so that 2n + 1, n2 + n + 1 and 3n2 – 3n + 3 are consecutive terms of an A .P.

    Solution

    Solution not provided.
    Ans. 1  or 2

    Question 82
    CBSEENMA10007651
    Wired Faculty App

    Determine the general term of an A .P. whose 5th term is 6 and 7th term is 12.

    Solution

    Solution not provided.
    Ans. 3n - 9

    Question 83
    CBSEENMA10007652
    Wired Faculty App

    The seventh term of an A .P. is 20 and 3rd term of an A .P. be 8. Find 32nd term and the general term.

    Solution

    Solution not provided.
    Ans. 95, 3n - 1

    Question 84
    CBSEENMA10007653
    Wired Faculty App

    The sum of three numbers in A .P. is -3 and the product is 35. Find the numbers.

    Solution

    Solution not provided.
    Ans. -7, -1, - 5,  or 5, - 1,  - 7,  .....

    Question 85
    CBSEENMA10007654
    Wired Faculty App

    The sum of three consecutive terms in an A .P. is 18 and their product is 192. Find the numbers.

    Solution

    Solution not provided.
    Ans. 4, 6, 8  or 8, 6, 4

    Question 87
    CBSEENMA10007656
    Wired Faculty App

    The first term of an A .P. is 5 and its 100th term is –292. Find the 50th term of this A .P.
    Solution

    Solution not provided.
    Ans. - 145

    Question 88
    CBSEENMA10007664
    Wired Faculty App

    The sum of the first n terms of an A .P. is 3n2 + 2n. Find the nth term.

    Solution

    Solution not provided.
    Ans. 6n - 1

    Question 90
    CBSEENMA10007670
    Wired Faculty App

    Find the sum of n terms of the sequence (an)where an = 5 – 6n.

    Solution

    Solution not provided.
    Ans. space 2 straight n space minus space 3 straight n squared

    Question 91
    CBSEENMA10007671
    Wired Faculty App

    If the sum of first n terms of an A .P. is given by Sn = 5n2 + 3n. Find the nth term of the A .P.

    Solution

    Solution not provided.
    Ans. 10n - 2

    Question 95
    CBSEENMA10007676
    Wired Faculty App

    The sum of first eight terms of an A .P. is 0 and its sixth term is 6. Find the first term and common difference.

    Solution

    Solution not provided.
    Ans. a = -14, d= 4

    Question 97
    CBSEENMA10007679
    Wired Faculty App

    Find the sums given below :

    (i) 101 + 99 + 97 +.........+ 47
    (ii) 5 + 8 + 11 +.........+ 185
    (iii) 2 + 4 + 6 +.........+ 200
    (iv) 3 + 11 + 19 +.........+ 803
    Solution

    Solution not provided.
    Ans. (i) - 336 (ii) 5795 (iii) 10100 (iv) 40703

    Question 99
    CBSEENMA10007683
    Wired Faculty App

    In an A .P.
    (i) given a = 5, d = 3, Tn = 50, find n and Sn.
    (ii) given a = 6, d = 4, Tn = 50 find n and Sn.
    (iii) given a = 7, d = 6, Tn = 43, find n and Sn.
    Solution

    Solution not provided.
    Ans.  (i) 16; 44,  (ii) 12; 336   (iii) 7; 175

    Question 100
    CBSEENMA10007684
    Wired Faculty App

    Find the sum of all odd numbers between 100 and 200.

    Solution

    Solution not provided.
    Ans.  7500

    Question 101
    CBSEENMA10007685
    Wired Faculty App

    Find the sum of all even integers between 101 and 599.

    Solution

    Solution not provided.
    Ans.  87150

    Question 102
    CBSEENMA10007686
    Wired Faculty App

    Find the sum of first natural numbers.

    Solution

    Solution not provided.
      space space space space space space space space Ans. space space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction

    Question 103
    CBSEENMA10007688
    Wired Faculty App

    Find the sum of all multiples of 5 lying between 101 and 999.

    Solution

    Solution not provided.
    Ans.  98450

    Question 104
    CBSEENMA10007689
    Wired Faculty App

    Find the sum of first 100 even natural numbers divisible by 5.
    Solution

    Solution not provided.
    Ans.  50500

    Question 109
    CBSEENMA10007696
    Wired Faculty App

    How many terms of the sequence 18, 16, 14,......should be taken so that their sum is zero.

    Solution

    Solution not provided.
    Ans.  19

    Question 111
    CBSEENMA10007698
    Wired Faculty App

    Find the sum of first 24 terms of the sequence whose nth term is giver. by   straight u subscript straight n equals 3 plus 2 over 3 straight n.

    Solution

    Solution not provided.
    Ans. 272

    Question 116
    CBSEENMA10007703
    Wired Faculty App

    Find the sum of all three digit numbers each of which leave the remainder 3 when divided by 5.

    Solution

    Solution not provided.
    Ans.  99090

    Question 118
    CBSEENMA10007706
    Wired Faculty App

    If the sum of first n terms of an A .P. is 3n2 – 2n, find the A .P. and its 19th term.

    Solution

    Solution not provided.
    Ans. 109

    Sponsor Area

    Question 120
    CBSEENMA10007708
    Wired Faculty App

    The nth term an of an A .P. is given by an = 5 – 2n Find the sum of first 25 terms of the A .P.

    Solution

    Solution not provided.
    Ans. - 525

    Question 121
    CBSEENMA10007709
    Wired Faculty App

    Find the number of terms of the A .P. 54, 51, 48,........so that their sum is 513.

    Solution

    Solution not provided.
    Ans. 18

    Question 122
    CBSEENMA10007710
    Wired Faculty App

    Find the number of terms of A .P. 64, 60, 56........so that their sum is 544.

    Solution

    Solution not provided.
    Ans. 16 or 17

    Question 124
    CBSEENMA10007712
    Wired Faculty App

    If the sum of n terms of an A .P. is given by Sn = 3n2 + 2n Find the nth term of the A .P.

    Solution

    Solution not provided.
    Ans. 6n - 1

    Question 126
    CBSEENMA10007715
    Wired Faculty App

    The 7th term of an A .P is 32 and its 13th term is 62. Find the A .P.

    Solution

    Solution not provided.
    Ans. 2, 7, 12, 17, ......

    Question 127
    CBSEENMA10007716
    Wired Faculty App

    Find the sum of the first 25 terms of an A .P. whose 4th term is given by tn= 2 – 3n.

    Solution

    Solution not provided.
    Ans.- 925

    Question 128
    CBSEENMA10007718
    Wired Faculty App

    The 7th terms of an A .P. is 20 and its 13th term is 32. Find the A .P.

    Solution

    Solution not provided.
    Ans. 8, 10, 12, 14,  .....

    Question 129
    CBSEENMA10007719
    Wired Faculty App

    The 7th term of an A .P. is –4 and its 13th term is –16. Find the A .P.

    Solution

    Solution not provided.
    Ans. 8, 4, 6, 2, 0, .....

    Question 130
    CBSEENMA10007720
    Wired Faculty App

    The 8th terms of an Arithmetic Progression (A .P.) is 37 and its 12th term is 57. Find the A .P.

    Solution

    Solution not provided.
    Ans. 2, 7, 12 , 17, ......

    Question 131
    CBSEENMA10007722
    Wired Faculty App

    Find the sum of the first 25 terms of an A .P. whose nth term is given by tn= 7 – 3n.

    Solution

    Solution not provided.
    Ans. -880

    Question 132
    CBSEENMA10007723
    Wired Faculty App

    Which term of the Arithmetic Progression 3, 10, 17........will be 84 more than its 13th term ?

    Solution

    Solution not provided.
    Ans. 25

    Question 133
    CBSEENMA10007724
    Wired Faculty App

    The 8th term of an Arithmetic Progression (A .P.) is –23 and its 12th term is 39. Find the A .P.

    Solution

    Solution not provided.
    Ans. 5, 1, -3, 7, ......

    Question 134
    CBSEENMA10007725
    Wired Faculty App

    The 8th term of an Arithmetic Progression (A .P.) is 32 and its 12th term is 52. Find the A .P.

    Solution

    Solution not provided.
    Ans. -3, 2, 7, 12, ...

    Question 135
    CBSEENMA10007727
    Wired Faculty App

    If the sum of first n terms of an A .P. given by Sn = 3n2+ 2n. find the nth term of the A .P.

    Solution

    Solution not provided.
    Ans. 6n -1

    Question 136
    CBSEENMA10007729
    Wired Faculty App

    If m times the mth term of an A.P. is equal to n times its nth term, find its (m + n)th term.

    Solution

    Solution not provided.
    Ans. 0

    Question 137
    CBSEENMA10007730
    Wired Faculty App

    If the sum of first n terms of an A .P. is given by Sn = 2n2 + 5n, find the nth term of the A .P.

    Solution

    Solution not provided.
    Ans. 4n + 3

    Question 138
    CBSEENMA10007731
    Wired Faculty App

    Find the sum of first 15 terms of an A .P. whose nth term is 9 – 5n.

    Solution

    Solution not provided.
    Ans. -465

    Question 139
    CBSEENMA10007733
    Wired Faculty App

     If the sum of first n terms of an A .P. is given by Sn = 5n2 + 3n, find the nth term at the A .P.

    Solution

    Solution not provided.
    Ans. 10n - 2

    Question 140
    CBSEENMA10007736
    Wired Faculty App

    Find the number of terms of the A .P. 54, 51, 48.......so that their sum is 513.

    Solution

    Solution not provided.
    Ans.  18 ro 19

    Question 141
    CBSEENMA10007738
    Wired Faculty App

    If the nth term of an A .P. is (2n + 1), find the sum of first n terms of the A .P.

    Solution

    Solution not provided.
    Ans.  n(n+2)

    Question 142
    CBSEENMA10007739
    Wired Faculty App

    Find the sum of all two digit odd positive numbers.

    Solution

    Solution not provided.
    Ans.  2475

    Question 143
    CBSEENMA10007740
    Wired Faculty App

    Find the sum of all multiples of 9 lying between 300 and 700.

    Solution

    Solution not provided.
    Ans.  21, 978

    Question 144
    CBSEENMA10007741
    Wired Faculty App

    Find the sum of all multiples of 7 lying between 500 and 800.

    Solution

    Solution not provided.
    Ans.  27993

    Question 145
    CBSEENMA10007742
    Wired Faculty App

    Which term of the A .P. 5, 9, 13,..........is 81 ? Also find the sum 5 + 9 + 13 +.........+ 81.

    Solution

    Solution not provided.
    Ans. 860

    Question 146
    CBSEENMA10007743
    Wired Faculty App

    Which term of the A .P. 5, 9, 13,..........is 81 ? Also find the sum 5 + 9 + 13 +.........+ 81.

    Solution

    Solution not provided.
    Ans. 31

    Question 147
    CBSEENMA10007745
    Wired Faculty App

    The sum of first n terms of an A .P. is given by (n2 + 5n), find the 25th term of the progression.

    Solution

    Solution not provided.
    Ans. 54

    Question 148
    CBSEENMA10007746
    Wired Faculty App

    Which term of the A .P. 6, 10, 14..........is 102 ? Also find the sum 6 + 10 + 14 +........... 102.

    Solution

    Solution not provided.
    Ans. 1350

    Question 149
    CBSEENMA10007747
    Wired Faculty App

    Find the sum of the first 51 terms of the A .P. whose 2nd term is 2 and nth term is 8.

    Solution

    Solution not provided.
    Ans. 3774

    Question 151
    CBSEENMA10007749
    Wired Faculty App

    The sum of the first n terms of an A .P. is given by Sn = 3n2 – 4n. Determine the A .P. and its 12th terms.

    Solution

    Solution not provided.
    Ans. -1, 5, 11, 17  and T12 = 65

    Question 152
    CBSEENMA10007751
    Wired Faculty App

    The sum of three numbers in A .P. is 3 and their product is -35. Find the numbers.

    Solution

    Solution not provided.
    Ans. -5, 1, 7 or 7, 1, -5

    Question 153
    CBSEENMA10007752
    Wired Faculty App

    Find the sum of the first 21 terms of the A .P. whose 2nd terms is 8 and 4th term is 14.
    Solution

    Solution not provided.
    Ans. 735

    Question 154
    CBSEENMA10007753
    Wired Faculty App

    The sum of the first n terms of an A .P. is given by Sn = 3n2 + 2n. Determine the A .P. and its 15th term.
    Solution

    Solution not provided.
    Ans. 5, 11, 17, 23;, T15 = 89

    Question 155
    CBSEENMA10007754
    Wired Faculty App

    The sum of the first n terms of an A .P. is given by Sn = 3n2 + 2n. Determine the A .P. and its 15th term.
    Solution

    Solution not provided.
    Ans. 3, 7, 11 or 11, 7, 3

    Question 156
    CBSEENMA10007755
    Wired Faculty App

    Find the sum of the first 25 terms of the A .P. whose 2nd term is 9 and 4th term is 21.

    Solution

    Solution not provided.
    Ans. 1875

    Question 158
    CBSEENMA10007758
    Wired Faculty App

    Find the sum of all the two digit natural numbers which are divisible by 4.

    Solution

    Solution not provided.
    Ans. 1188

    Question 159
    CBSEENMA10007759
    Wired Faculty App

    Find the sum of all natural numbers between 100 and 200 which are divisible by 4.

    Solution

    Solution not provided.
    Ans. 3600

    Question 160
    CBSEENMA10007760
    Wired Faculty App

    Find the sum of all the natural numbers less than 200 which are divisible by 5.

    Solution

    Solution not provided.
    Ans. 3900

    Question 162
    CBSEENMA10007762
    Wired Faculty App

    Find (he sum of all the natural numbers less than 100 which are divisible by 6.

    Solution

    Solution not provided.
    Ans. 816

    Question 163
    CBSEENMA10007763
    Wired Faculty App

    Find the sum of all the natural numbers between 100 and 500 which are divisible by 8.

    Solution

    Solution not provided.
    Ans. 15, 000

    Question 164
    CBSEENMA10007764
    Wired Faculty App

    Find the sum of all the natural numbers between 200 and 300 which are divisible by 4.

    Solution

    Solution not provided.
    Ans. 6000

    Question 165
    CBSEENMA10007765
    Wired Faculty App

    If the 10th term of an A .P. is 52 and the 17th term is 20 more than the 13th term, find the A .P.

    Solution

    Solution not provided.
    Ans. 7, 12, 17, 22, .....

    Question 166
    CBSEENMA10007766
    Wired Faculty App

    Using A .P., find the sum of all 3 digit natural number which are multiples of 7.

    Solution

    Solution not provided.
    Ans. 70336

    Question 167
    CBSEENMA10007767
    Wired Faculty App

    If the 8th term of an A .P. is 31 and the 15th term is 16 more than the 11th term, find the A .P.

    Solution

    Solution not provided.
    Ans. 3, 7, 11, 15, .....

    Question 168
    CBSEENMA10007768
    Wired Faculty App

    Find the sum of the following : 25 + 28 + 31 +.........+ 100.

    Solution

    Solution not provided.
    Ans. 1625

    Question 169
    CBSEENMA10007769
    Wired Faculty App

    Find the sum of all 3 digit natural numbers which are divisible by 13.

    Solution

    Solution not provided.
    Ans. 37674

    Question 170
    CBSEENMA10007770
    Wired Faculty App

    Which term of the arithmetic progression 5, 15, 25...........will be 130 more than its 31st term?

    Solution

    Solution not provided.
    Ans. 44th term

    Question 171
    CBSEENMA10007771
    Wired Faculty App

    In an A .P, the sum of first n terms is  fraction numerator 5 straight n squared over denominator 2 end fraction plus fraction numerator 3 straight n over denominator 2 end fraction find the 20th term.

    Solution

    Solution not provided.
    Ans. 99

    Question 172
    CBSEENMA10007772
    Wired Faculty App

    Which term of the arithmetic progression 8. 14. 20, 26..........will be 72 more than its 41st term?

    Solution

    Solution not provided.
    Ans. 53rd term

    Question 174
    CBSEENMA10007774
    Wired Faculty App

    In an A .P., the sum of first n terms is fraction numerator 3 straight n squared over denominator 2 end fraction plus fraction numerator 13 straight n over denominator 2 end fraction find its 25th term.
    Solution

    Solution not provided.
    Ans. 80

    Question 175
    CBSEENMA10007775
    Wired Faculty App

    The 10th term of an A .P. is 47 and its 15th term is 72. Find the A .P.

    Solution

    Solution not provided.
    Ans. 2, 7, 12, ......58

    Question 177
    CBSEENMA10007777
    Wired Faculty App

    The nth term (Tn) of an A .P. s given by Tn = 7n + 1. Find the sum of the first 30 terms of A .P.

    Solution

    Solution not provided.
    Ans. 3285

    Question 178
    CBSEENMA10007778
    Wired Faculty App

    The 10th term of an A .P. is 57 and its 15th term is 87. Find A .P.

    Solution

    Solution not provided.
    Ans. 3, 9, 15, 21, .........

    Question 179
    CBSEENMA10007779
    Wired Faculty App

    Which term of the A .P. 3, 15, 27, 39.....will be 132 more than its 54th term.

    Solution

    Solution not provided.
    Ans. 65th term

    Question 180
    CBSEENMA10007780
    Wired Faculty App

    Find the sum of 1st 18th term of an A .P. whose nth term is 3 - 3n.

    Solution

    Solution not provided.
    Ans. - 288

    Question 181
    CBSEENMA10007781
    Wired Faculty App

    Which term of the A .P. 3. 15, 27, 39...........will be 132 more than its 60th term.

    Solution

    Solution not provided.
    Ans. 60th term

    Question 182
    CBSEENMA10007782
    Wired Faculty App

    In an A .P. the sum of its first n term is n2 + 2n. Find its 18th term.

    Solution

    Solution not provided.
    Ans. 37

    Question 183
    CBSEENMA10007783
    Wired Faculty App

    The third terms of an A .P. is three and 11th term is – 21. Find its Ist term and common difference.

    Solution

    Solution not provided.
    Ans. a = 9, b = -3

    Question 184
    CBSEENMA10007784
    Wired Faculty App

    The nth term of an A .P. is 7 - 4/7. Find its common difference.

    Solution

    Solution not provided.
    Ans. - 4

    Question 185
    CBSEENMA10007785
    Wired Faculty App

    The nth term of an A .P. is 6n + 2. Find its common difference.

    Solution

    Solution not provided.
    Ans. 67

    Question 186
    CBSEENMA10007786
    Wired Faculty App

    Write the next term of the A .P.

    square root of 8 comma space square root of 18 comma square root of 32 comma space...........

    Solution

    Solution not provided.
    Ans. space square root of 50

    Question 187
    CBSEENMA10007787
    Wired Faculty App

    Write the next term of the A .P.
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    Solution

    Solution not provided.
    Ans. space space square root of 32

    Question 188
    CBSEENMA10007788
    Wired Faculty App

    The first term of an A .P. is P and its common difference is q. Find its 10th term.

    Solution

    Solution not provided.
    Ans. P + 9q

    Question 189
    CBSEENMA10007789
    Wired Faculty App

    The sum of n terms of an A .P. is 3n2 + 5n. Find the A .P. Hence, find its 16th term.

    Solution

    Solution not provided.
    Ans. 98

    Question 191
    CBSEENMA10007791
    Wired Faculty App

    Find the sum of all two digit natural numbers which when divided by 3 yield 1 as remainder. 1605

    Solution

    Solution not provided.
    Ans. 1605

    Question 192
    CBSEENMA10007792
    Wired Faculty App

    For what value of n are the nth terms of two A .P.’s 63, 65, 67 .......and 3, 10, 17,.... equal.

    Solution

    Solution not provided.
    Ans. 13

    Question 196
    CBSEENMA10007796
    Wired Faculty App

    In an A .P., the term is 22, nth term is –11 and sum to first n terms is 66. Find n and d, the common difference.

    Solution

    Solution not provided.
    Ans. n = 12, d = -3

    Question 197
    CBSEENMA10007797
    Wired Faculty App

    Find the sum of all two digit natural numbers which common divided by 7 yield 1 as remainders.

    Solution

    Solution not provided.
    Ans.741

    Question 198
    CBSEENMA10007798
    Wired Faculty App

    Write the common difference of an A .P. whose nth term is 3n + 5.

    Solution

    Solution not provided.
    Ans. 3

    Question 199
    CBSEENMA10007799
    Wired Faculty App

    Find the middle term of the A .P. : 10, 7, 4...... (–62).

    Solution

    Solution not provided.
    Ans. -26

    Question 200
    CBSEENMA10007800
    Wired Faculty App

    For what value of p are 2p + 1, 13, 5p – 3, three consecutive terms of an A .P.

    Solution

    Solution not provided.
    Ans. p = 4

    Question 201
    CBSEENMA10007801
    Wired Faculty App

    If Sn, the sum of first n terms of an A .P. is given by Sn = 3n2 – 4n, then find its nth term.

    Solution

    Solution not provided.
    Ans. 6n - 7

    Question 202
    CBSEENMA10007802
    Wired Faculty App

    If Sn, the sum of first n terms of an A .P. is given by Sn = Sn2 + 3n, then find its nth term.

    Solution

    Solution not provided.
    Ans. 10n - 2

    Question 203
    CBSEENMA10007803
    Wired Faculty App

    The sum of 4th and 8th terms of an A .P. is 24 and sum of 6th and 10th terms is 44. Find A .P.

    Solution

    Solution not provided.
    Ans. -13, -8, -3, 2, .....

    Question 204
    CBSEENMA10007804
    Wired Faculty App

    The sum of 5th and 9th terms of an A .P. is 72 and the sum of 7th and 12th terms is 97. Find the A .P.

    Solution

    Solution not provided.
    Ans. 6, 11, 16,  21, .....

    Question 205
    CBSEENMA10007805
    Wired Faculty App

    If 4/5, a, 2 are three consecutive terms of an A .P., then find the value of a.

    Solution

    Solution not provided.
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    Question 206
    CBSEENMA10007806
    Wired Faculty App

    Which term of the A .P. 3, 15, 27, 39,.....will be 120 more than its 21 st term.

    Solution

    Solution not provided.
    Ans. 31st  term

    Question 208
    CBSEENMA10007808
    Wired Faculty App

    For what value of k, are the numbers x, 2x + k and 3x + 6 three consecutive terms of an A .P.?

    Solution

    Solution not provided.
    Ans. k = 3

    Question 209
    CBSEENMA10007809
    Wired Faculty App

    Which term of AP; 121, 117, 113,.....is its first negative terms ?
    (a) 31 (b) 33 (c) 32 (d) 34
    Solution

    Solution no provided.
    Ans. (b)

    Question 211
    CBSEENMA10007811
    Wired Faculty App

    The 8th term of 117, 104, 91, 78,.........

    (a) 25 (b) 26 (c) 27 (d) 28
    Solution

    Solution not proovided.
    Ans. (b)

    Question 212
    CBSEENMA10007812
    Wired Faculty App

    The value of ‘A’ so that  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> arc three consecutive terms of an AP.

    left parenthesis straight a right parenthesis space 16 over 3 space space space space left parenthesis straight b right parenthesis space 16 over 15 space space space space space left parenthesis straight c right parenthesis space 16 over 33 space space space space space left parenthesis straight d right parenthesis space space 16 over 35

    Solution

    Solution not proovided.
    Ans. By definition of A. P.

    Question 218
    CBSEENMA10007818
    Wired Faculty App

    The next term of the AP.   

    square root of 8 comma space square root of 18 comma space square root of 32 comma space........ space is
    Solution

    Solution not provided.
    Ans. (b)

    Question 219
    CBSEENMA10007819
    Wired Faculty App

    The 7th term of an AP is 32 and 13th term is 62. The value of common difference is
    (a) 4 (b) 5 (c) 6 (d) 7
    Solution
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    On solving (i) and (ii) : d = 5,   
    Ans. (b)
    Question 221
    CBSEENMA10007821
    Wired Faculty App

    The 10th term from the end of the A .P., 8,10, 12,........ 126 is
    (a) 107 (b) 106 (c) 108 (d) 109
    Solution

    Solution not provided.
    Ans. (c)

    Question 223
    CBSEENMA10007823
    Wired Faculty App

    The sum of all two digit positive numbers divisible by 3 is
    (a) 1664 (b) 1665 (c) 1666 (d) 1600
    Solution

    Solution not provided.
    Ans. (b)

    Question 225
    CBSEENMA10007825
    Wired Faculty App

    If a6 – a5 is 200 then the value of a7 – a6 of the same A .P. will be
    (a) 100 (b) 400 (c) 200 (d) 300
    Solution

    Solution not provided.
    Ans. (c) [Difference between any two consecutive numbers is always same]

    Question 226
    CBSEENMA10007826
    Wired Faculty App

     Find the sum of three numbers is an A .P. is 27 and their product is 405. The numbers are : (a) (3,9,15) (b) (15, 9, 3) (c) both (a) and (b) (d) (3, 9, 12)
    Solution

    Let the three numbers are a - d, a and a + d
    space space space space space space space therefore (a - d) +a + (a + d) = 27 rightwards double arrow a = 9, Also, (a - d) (a) (a + d) = 405 
    Ans (c)

    Question 228
    CBSEENMA10007828
    Wired Faculty App

    In which of the following situational, does the list of numbers involved make an arithmetic progression, and why?

    The cost of digging a well for the first metre is Rs. 150 and rises by Rs. 50 for each succeeding metre.

    Solution

    Cost of digging the well after 1 metre of digging of Rs. 150 = a4
    Cost of digging the well after 2 metres of digging
    = Rs. 150 + Rs. 50
    = Rs. 200 = a2
    Cost of digging the well after 3 metres of digging
    = Rs. 150 + Rs. 50

    = Rs. 2a = a3
    Cost of digging the well after 4 metres of digging
    = Rs. 200 + Rs. 50
    = Rs. 250 = a4
    and so on.
    a2 – a4 = Rs. 200 – Rs. 150 = Rs. 50
    a3 – a2 = Rs. 250 – Rs. 200 = Rs. 50
    a4 – a3 = Rs. 350 – Rs. 250 = Rs. 50
    i.e., ak +1 – ak is the same every time.
    So this list of numbers forms an A .P. with the first term a Rs. 150 and the common difference d = Rs. 50.

    Question 229
    CBSEENMA10007829
    Wired Faculty App

    Write first four terms the A .P. when the first term ‘a’ and the common difference ‘d’ are given as follows :

    a – 10, d = 10

    Solution

    a = 10, d = 10
    First term = a = 10
    Second term = 18 + d = 10+ 10 = 20
    Third term = 20 + d = 20 + 10 = 30
    Fourth term = 30 + d = 30 + 10 = 40
    Hence, first four terms of the given A .P. are 10, 20, 30, 40.

    Question 230
    CBSEENMA10007830
    Wired Faculty App

    Write first four terms the A .P. when the first term ‘a’ and the common difference ‘d’ are given as follows :

    a = –2, d = 0

    Solution

    a = –2, d = 0
    First term = a = – 2
    Second term = –2 + d = –2 + 0 = –2
    Third term = –2 + d = –2 + 0 = –2
    Fourth term = –2 + d = –2 + 0 = –2
    Hence first, first four terms of the given A .P. are -2, -3, -2 -2.

    Question 231
    CBSEENMA10007833
    Wired Faculty App

    The 17th term of an A .P. exceeds the 10th term by 7. Find the common difference.

    Solution

    We have
    a17 = a + 16d
    and a10 = a + 9d
    According to the given question,
    a17 – a10 = 7
    ⇒ (a + 16d) – (a + 9d) = 7
    ⇒ a + 16d – a – 9d = 7
    ⇒ 7d – 7
    ⇒ d = 1
    Hence, common difference is 1.

    Question 232
    CBSEENMA10007834
    Wired Faculty App

    Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 100th terms?

    Solution

    Let the common difference of both the
    A.P.’s = d
    and First term of Ist A.P. = a
    also first term of 2nd A .P. = b
    Then, 200th term of 1st A .P. = a + 99d
    and 100th term of IInd A .P. = b + 99d
    Difference between their 100th term = 100
    (given)
    ⇒ (a + 99d) – (b + 99d) = 100
    ⇒ a – b = 100 ...(i)
    Now, a1000 of Ist A.P. = a + 999d
    and a1000 of Ilnd A .P. = b + 999d
    Hence, difference between their 1000th terms
    = (a + 999d) (b + 999d) = a – b
    = 100. [Using (i)]

    Question 233
    CBSEENMA10007851
    Wired Faculty App

    In an A.P. :
    Given a = 2, d = 8, Sn = 90, find ‘n’ and an
    Solution

    Here,    a = 2, d = 8, Sn = 90
    We know that,
                Sn = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> [2 x 2 + (n -1)d]
    rightwards double arrow     180  =n[2 x 2 (n - 1) x 8]
    rightwards double arrow     180 = n(8n - 4)
    rightwards double arrow     180 = 8n2 - 4n
    rightwards double arrow     8n2 - 4n - 180 = 0
    rightwards double arrow     2n2 - n - 45 = 0     [Dividing by 4]
    rightwards double arrow    2n(n - 5) (2n + 9) = 0
    rightwards double arrow    n - 5 = 0  or 2n + 9 = 0
    rightwards double arrow    n = 5 or n = fraction numerator negative 9 over denominator 2 end fraction
    But number of terms cannot be (- ve)
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Now,   a5 = a+ 4d
                  = 2 + 4 + 8
    Hence,     n = 5      
    an = 34

    Question 234
    CBSEENMA10007854
    Wired Faculty App

    In an A.P. 
    Given a = 8, an = 62, Sn = 210, find ‘n’ and ‘d’.
    Solution

    Here a = 8
    straight a subscript straight n space equals space 62 end subscript
    Sn = 210
    We know that, an = a + (n - 1)d
    rightwards double arrow   62 - 8 = (n - 1)d
    rightwards double arrow   54 = (n -d)d                   ...(i)
    And,  Sn = n over 2 '[ 2a - (n - 1) d
    rightwards double arrow  210 = n over 2[ 2 x 8 + (n - 1) d]
    rightwards double arrow  420 = n[16 +(n - 1)d] .......(ii)
    From (i) and (ii), we get
    420  = n[ 16 + 54]
    rightwards double arrow   420  = 70n
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Putting this value of n in (i) we get
    54 = (6 -1) x d
    rightwards double arrow     5d = 54
    rightwards double arrow     d = 54 over 5
    Hence,  n = 6 and d = space space 54 over 5.

    Question 235
    CBSEENMA10007856
    Wired Faculty App

    In an A.P. 
    Given an = 4, d = 2, Sn = –14, find ‘n’ and ‘a’.
    Solution

    Here,  an = 4,     d = 2
    Sn = –14
    We know that,  
        an  = a + (n -1)d
    rightwards double arrow  4 = a + (n -1) x 2
    rightwards double arrow  4 = a + 2n - 2
    rightwards double arrow  a + 2n = 6 rightwards double arrow a = 6 - 2 n    ....(i)
    And.     Sn = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>  [2a+ (n - 1)d]
    rightwards double arrow           - 14 = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    rightwards double arrow            - 28 = n[2a + 2 (n -1) ]
    rightwards double arrow            - 28 = 2n [a + (n -1)]
    rightwards double arrow            - 14 = n[a + (n - 1)]      ......(ii)

    Putting the value of (i) in (ii), we get,
    – 14 = n [6 – 2n + n – 1]
    ⇒ – 14 = n[5 – n]
    ⇒ –14 = 5n – n2
    ⇒ n2 –5n – 14 = 0
    ⇒ (n – 7) (n + 2) = 0
    ⇒ n – 7 = 0 or n + 2 = 0
    ⇒ n = 7 or n = –2
    But n cannot be –ve. ∴ n = 7.
    Putting this value of V in equation (i), we get
    a + 2 x 7 = 6
    ⇒ a = 6 – 14 = – 8
    Hence, n = 7 and a = –8


    Question 236
    CBSEENMA10007859
    Wired Faculty App

    In an A.P. 
    Given a = 3, n = 8, S = 192, find ‘d’.
    Solution

    Here a = 3, n = 8
    S = 192
    We know that,  
           straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket
    therefore space space space space 192 space equals space 8 over 2 left square bracket 2 cross times 3 plus left parenthesis 8 minus 1 right parenthesis straight d right square bracket rightwards double arrow space space space space 192 over 4 equals 6 space plus space 7 straight d
    rightwards double arrow   6 +7d = 48
    rightwards double arrow space space space space straight d space equals space 42 over 7 equals 6
    Hence, d = 6 Ans.

    Question 237
    CBSEENMA10007862
    Wired Faculty App

    In an A.P. 
    Given l = 28, S = 144, n = 9, find ‘a’.
    Solution

    Here, l = 28, S = 144
    n = 9
    l = a + (n – 1)d
    28 = a + (9 – 1)d
    ⇒ 28 = a + 8d ...(i)
    And,
          <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    rightwards double arrow space space space space space space 288 over 9 equals 2 left parenthesis straight a space plus space 4 straight d right parenthesis
    rightwards double arrow    32 = 2(a + 4d)
    rightwards double arrow    a + 4d = 16                                  .....(ii)
    Solving (i) and (ii) we get,
          a = = 32 -28 = 4
    Hence ,    a   =  4

    Question 238
    CBSEENMA10007866
    Wired Faculty App

    How many terms of the A.P. 9, 17, 25, ... must be taken to give a sum of 636? 
    Solution

    But number of terms cannot be –ve. ∴ n = 12.
    Hence, the number of terms is 12.

    Question 239
    CBSEENMA10007867
    Wired Faculty App

    The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
    Solution

    Here a = 5, l = 45, S = 400
    We know that,

     
    Question 240
    CBSEENMA10007869
    Wired Faculty App

    The first and the last term of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum ?

    Solution

    Here,     a = 17,  l = 350,   d = 9
    We know that, 
                            l = a + (n - 1)d
    rightwards double arrow    350 = 17 + (n -1) x 9
    rightwards double arrow   350 -17 = 9 x (n - 1)
    rightwards double arrow    333 = 9 x (n -1)
    rightwards double arrow    n - 1 = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>     n = 37 + 1 = 38
    Hence,    there are 28 terms.
    because space space space space space straight S subscript straight n space equals space straight n over 2 left parenthesis straight a plus 1 right parenthesis therefore space space space space space space straight S subscript 38 equals space 38 over 2 left parenthesis 17 plus 350 right parenthesis
    = 19 x 367 = 6973.

    Question 241
    CBSEENMA10007874
    Wired Faculty App

    Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
    Solution

    Here n = 51, a2 = 14, a3 = 18
    d = a3 – a2 = 18 – 14 = 4 a2 = a + d
    ⇒ 14 = a + 4
    ⇒ a = 14 – 4 = 10

    We know that,
         straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d therefore space straight S subscript 51 equals 51 over 2 left square bracket 2 cross times 10 plus left parenthesis 51 minus 1 right parenthesis cross times 4 right square bracket equals 51 over 2 left square bracket 20 plus 50 cross times 4 right square bracket equals 51 over 2 cross times left parenthesis 20 plus 200 right parenthesis equals 51 over 2 cross times 220 equals 51 cross times 110 equals 5610.

    Question 242
    CBSEENMA10007875
    Wired Faculty App

    If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
    Solution
    Here S7 = 49, S17 = 289 Let ‘a’ be the first term and ‘d’ the common difference.
    Question 243
    CBSEENMA10007877
    Wired Faculty App

    Show that a1 , a2 , . . ., an, . . . form an AP where an is defined as below :
    an = 3 + 4n
    Solution

    It is given that,
             an = 3 + 4n
    So,    a1 = 3 + 4(1) = 3 + 4 = 7
             a2  = 3 + 4 (2) = 3 + 8 = 11
    and    a3 = 3 + 4(3) = 3 + 12 = 15
    Now, er have following numbers :
        7, 11, 15, ...........
    Since the difference between each pair of consecutive terms are constant
    So, the given number forms an A. P.
    Here, a = 7, d = 4,    n  = 15
    We know that, 
    straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket So comma space space space straight S subscript 25 equals 25 over 2 left square bracket 2 space straight x space 7 space plus space left parenthesis 25 space minus space 1 right parenthesis space 4 right square bracket space space space space space space space space space space space space space space equals 25 over 2 left square bracket 14 plus 96 right square bracket space equals space 25 over 2 cross times 110 space space space space space space space space space space space space space space space equals space 25 cross times 55 space equals space 1375

    Question 244
    CBSEENMA10007880
    Wired Faculty App

    Show that a1 , a2 , . . ., an, . . . form an AP where an is defined as below :
    an = 9 – 5n
    Solution

    It is given that :
    an = 9 – 5n
    a1 = 9 – 5(1) = 9 – 5 = 4
    a2 = 9 – 5 (2) = 9 – 10 = –1
    and a3 = 9 – 5(3) = 9 – 15 = –6
    Now, we have following numbers :
    4, –1,–6, –11,.........
    Since, the difference between each pair of consecutive terms are constant.
    So, the given numbers form an A.P.
    Here, a = 4, d = –5, n = 15
          We space know comma space space space space space space straight S subscript straight n space equals straight n over 2 left square bracket 2 straight a space plus left parenthesis space straight n minus 1 right parenthesis straight d right square bracket space space space space space space space space space space space space space space space space space space So comma space space straight S subscript 15 equals 15 over 2 left square bracket 2 cross times 4 plus 915 minus 109 minus 50 right square bracket space space space space space space space space space space space space space space space space space equals 15 over 2 left square bracket 8 minus 70 right square bracket equals 15 over 2 cross times negative 62 space space space space space space space space space space space space space space space space space equals 15 cross times negative 32 equals negative 480

    Question 245
    CBSEENMA10007882
    Wired Faculty App

    If the sum of the first n terms of an AP is 4n – n2 , what is the first term (that is S1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
    Solution

    We have given that
    Sum of the first n terms = 4n – n2
    ⇒ Sn = 4n – n2
    Put, n = 1, then
    S1 = 4(1) – (1)2 = 4 – 1 = 3
    ⇒ d1 = 3
    Hence, the first term is
    Put n = 2, then
    S2 = 4(2) – (2)2
    = 8 – 4 = 4
    Hence, the sum of two terms is 4.
    Now, second term = S2 – S1 = 4 – 3 = 1
    Put n = 3, then
    S3 = 4(3) – (3)2
    = 12 – 9 = 3
    Third term = S3 – S2 = 3 – 4 = –1
    Put n = 9, 10
    S9= 4(9) – (9)2
    = 36 – 81 = –45
    S10= 4(10) – (10)2
    = 40 –100 = –60
    ∴ Tenth term = S10 – S9
    = – 60 – (–45)
    = – 60 + 45 = –15
    Now, Sn –1 = 4(n – 1) – (n – 1)2
    = (4n – 4) – (n2 + 1 – 2n)
    = 4n – 4 – n2 – 1 + 2n
    = 6n – n2 – 5
    ∴ nth term = Sn – Sn – 1
    = (4n – n2) – (6n – n2 – 5)
    = 4n – n2 – 6n + n2 + 5
    = 5 –2n.

    Question 246
    CBSEENMA10007885
    Wired Faculty App

    Find the sum of the first 40 positive integers divisible by 6.
    Solution

    The A .P.’s of the first 40 positive integers be 6. 12, 18, 24,......., 40 terms.
    Here, a = 6, d = 6, n = 40
    We space know space that comma space space straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight g right square bracket So comma space space space space space space space straight S subscript 49 equals 40 over 2 left square bracket 2 cross times 6 plus left parenthesis 40 minus 1 right parenthesis 6 right square bracket space space space space space space space space space space space space space space space space space equals 20 left square bracket 12 plus 39 cross times 6 right square bracket space space space space space space space space space space space space space space space space space equals 20 left parenthesis 12 plus 2340 space space space space space space space space space space space space space space space space space equals space 20 cross times 246 space space space space space space space space space space space space space space space space space equals space 4920.

    Question 247
    CBSEENMA10007887
    Wired Faculty App

    Find the sum of the first 15 multiples of 8.
    Solution

    Multiples of 8 are : 8, 16, 24, 32.....which form an A .P.
    It is given that, a = 8, d =16 – 8 = 8, here n = 15
    We know that,     straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket rightwards double arrow space straight S subscript straight n equals 15 over 2 left square bracket 2 cross times 8 plus left parenthesis 15 minus 1 right parenthesis cross times 8 right square bracket space space space space space space space space equals space 15 over 2 left square bracket 16 plus 14 cross times 8 right square bracket space space space space space space space space equals 15 over 2 left square bracket 16 plus 112 right square bracket space space space space space space space space equals 15 over 2 cross times 128 equals 15 cross times 64 equals 960
    Hence, the sum of first 15 multiples of 8 = 960.

    Question 248
    CBSEENMA10007889
    Wired Faculty App

    Find the sum of the odd numbers between 0 and 50.
    Solution

    Odd numbers between 0 and 50 are
    1, 3, 5, 7.....which form an A .P.
    With a = 1, d = 3 – 1 = 2, an = 49 We know that,

    Question 250
    CBSEENMA10007893
    Wired Faculty App

    A sum of Rs. 280 is to be used to award four prizes. If each prize after the first is Rs. 20 less than the next most valuable one, find the value of each of the prizes.

    Solution

    Since each prize is Rs. 20 less than its preceding prize, therefore, the values of the seven successive cash prizes will form an A .P.
    Let the first prize be Rs. a. Then, Successive prize = Rs. (a – 20), Rs. (a – 40),
    Rs. (a – 60)..........Here, first term = a
    Common difference = (a – 20) – a = –20 n = 7 and Sn = 700
    We space know space that comma space space straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n plus negative 1 right parenthesis straight d right square bracket rightwards double arrow space space space space space space space space space space space 700 equals 7 over 2 left square bracket 2 straight a plus left parenthesis 7 minus 1 right parenthesis left parenthesis negative 20 right parenthesis right square bracket rightwards double arrow space space space space space space space space space space 700 equals 7 over 2 left square bracket 2 straight a plus left parenthesis 6 right parenthesis left parenthesis negative 20 right parenthesis right square bracket rightwards double arrow space space space space space space space space space space 1400 space equals space 7 left square bracket 2 straight a minus 120 right square bracket rightwards double arrow space space space space space space space space space space space space 200 equals 2 straight a minus 120 rightwards double arrow space space space space space space space space space space space space space space 2 straight a equals 200 plus 120 rightwards double arrow space space space space space space space space space space space space space space space 2 straight a space equals 320 rightwards double arrow straight a space equals space 160

    Hence value of first prize = Rs. 160, and successive prizes are
    Rs. (a – 20), Rs. (a – 40), Rs. (a – 60) ...
    i.e. Rs. (160 – 20), Rs. (160 – 40), Rs. (160 – 60)...
    i.e. Rs. 140, Rs. 120, Rs. 100 ...

    Question 251
    CBSEENMA10007896
    Wired Faculty App

    In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
    Solution

    Three sections of class I will plant = 1 x 3 = 3
    Three sections of class II will plant = 2 x 3 = 6
    Three sections of class III will plant = 3 x 3 = 9 and so on.
    Three sections of class XII will plant = 12 x 3 = 36
    So, we get an A .P. 3, 6, 9, ..., 36.
    Here,      a =  3
                  d = 6 - 3 = 3
                  an = 36
    We know that,    an = a +(n -1) d
    rightwards double arrow        36 = 3 + (n - 1)d
    rightwards double arrow      36 = 3 + (n - 1) x 3
    rightwards double arrow      36 - 3 = 3 x (n - 1)
    rightwards double arrow      33 = 3 x (n - 1)
    rightwards double arrow space space space straight n minus 1 equals 33 over 3 equals 11 rightwards double arrow space space space space straight n space equals space 11 plus 1 equals 12 Now comma space space space space straight S subscript straight n equals straight n over 2 left parenthesis straight a subscript 1 plus straight a subscript straight n right parenthesis therefore space space space space space space space space space straight S subscript 12 space equals 12 over 2 left parenthesis 3 plus 36 right parenthesis space space space space space space space space space space space space space space equals space 6 cross times 39 equals 234
    Hence, the number of trees planted by the students = 234. Ans.

    Question 252
    CBSEENMA10007904
    Wired Faculty App

    A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles?  open parentheses Take space straight pi space equals space 22 over 7 close parentheses
    Solution

    We know that,
    Length of semicircle = πr
    l1, = π x 0.5 cm
    l2 = π x 1.0 cm
    3 = π x 1.5 cm
    And, l4 = π x 2.0 cm and so on.
    Let l be the total length of such a spiral made up of thirteen consecutive semi-circles.

    Hence, total length of such a spiral made up of thirteen consecutive semi-circles = 143 cm. Ans.

    Question 253
    CBSEENMA10007907
    Wired Faculty App

    00 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig. 5.2). In how many rows the 200 logs are placed and how many logs are in the top row?


    Fig. 5.2.

    Solution

    In the bottom row = 20 logs
    In the next row = 19 logs
    In the row next to it = 18 logs and so on.
    Thus, 20, 19, 18,... form an A .P. with
    a1 = 20, a2 = 19, a3 = 18
    Here d = 19 – 20 = (–1)
    and Sn = 200

    Hence, the required number of rows are 16 and in the top row there are 5 logs.

    Question 254
    CBSEENMA10007914
    Wired Faculty App

    In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see the given Fig.)


    A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
    Solution

    To pick up the first potato the distance run = 2 x 5 m
    To pick up the second potato the distance run = 2 x (5 + 3) m
    To pick up the third potato the distance run = 2 x (5 + 6) m and so on.
    Total distance the competitor has to run = 2 x 5 + 2 x (5 + 3) + 2 x (5 + 6) + ... upto 10 terms
    = 10 + 16 + 22 + .... which is an A .P.
    Here, a = 10, d = 16 – 10 = 6 and n = 10
    therefore space straight S subscript 10 equals 10 over 2 left square bracket 2 cross times 10 plus ? left parenthesis 10 minus 1 right parenthesis 6 right square bracket space open square brackets because space Sn equals straight n over 2 left square bracket 2 straight a left parenthesis straight n minus 1 right parenthesis straight d right square bracket close square brackets equals 5 space straight x space left square bracket 20 plus 54 right square bracket equals 5 space straight x space 74 space equals space 370
    Here the total istance the compatitar has to sum = 370 m.

    Tips: -

    [Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]

    Question 255
    CBSEENMA10007919
    Wired Faculty App

    Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find n for an < 0]
    Solution

    he given A .P. is 121, 117, 113, ... Here, a = 121
    d = 117 – 121 = –4 Let the nth term of the A.P. be the first negative term. Then,

    Least integral value of n = 32. Hence, 32nd term of the given A .P. is the first negative term 9.

    Question 256
    CBSEENMA10007921
    Wired Faculty App

    The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
    Solution

    Let the first term and the common difference of the A .P. be a and d respectively.
    According to the question,
    Third term + seventh term = 6

    Question 257
    CBSEENMA10007922
    Wired Faculty App

    A ladder has rungs 25 cm apart. (see the given Fig.). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2 1 half straight m  apart, what is the length of the wood required for the rungs?
    Solution

    Number of rungs (n) = 250 over 25 equals 10
                                   open square brackets 2 1 half straight m equals 250 space cm close square brackets

    Hence, there are 10 rungs.
    The length of the wood required for the rungs (S10).

    straight S subscript 11 equals 10 over 2 left parenthesis 45 plus 25 right parenthesis space equals space 350 space cm.
    Question 258
    CBSEENMA10007924
    Wired Faculty App

    The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
    Solution

    The consecutive numbers on the houses of a row are 1, 2, 3, ... 49.
    Clearly this list of number forming an A .P.
    Here, a = 1,d = 2 – 1 = 1
    According to the question,
    straight S subscript straight x minus 1 end subscript equals straight S subscript 49 minus straight S subscript straight x rightwards double arrow space space space space fraction numerator straight x minus 1 over denominator 2 end fraction left square bracket 2 straight a plus left parenthesis straight x minus 1 minus 1 right parenthesis straight d right square bracket equals 49 over 2 left square bracket 2 straight a plus left parenthesis 49 minus 1 right parenthesis straight d right square bracket minus straight x over 2 left square bracket 2 straight a plus left parenthesis straight x minus 1 right parenthesis 1 right square bracket space space space space space space space space space space space space space space space space space space space open square brackets because space straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket close square brackets rightwards double arrow space fraction numerator straight x minus 1 over denominator 2 end fraction left square bracket 2 left parenthesis 1 right parenthesis plus left parenthesis straight x minus 2 right parenthesis left parenthesis 1 right parenthesis right square bracket rightwards double arrow space space 49 over 2 left square bracket 2 left parenthesis 1 right parenthesis plus left parenthesis 48 right parenthesis left parenthesis 1 right parenthesis right square bracket minus straight x over 2 left square bracket 2 left parenthesis 1 right parenthesis plus left parenthesis straight x minus 1 right parenthesis straight d right square bracket rightwards double arrow space space space fraction numerator straight x minus 1 over denominator 2 end fraction left square bracket straight x right square bracket equals 1225 minus fraction numerator straight x left parenthesis straight x plus 1 right parenthesis over denominator 2 end fraction rightwards double arrow space space space space fraction numerator left parenthesis straight x minus 1 right parenthesis left parenthesis straight x right parenthesis over denominator 2 end fraction plus fraction numerator straight x left parenthesis straight x plus 1 right parenthesis over denominator 2 end fraction equals 1225 rightwards double arrow space space space space space straight x over 2 left parenthesis straight x minus 1 plus straight x plus 1 right parenthesis equals 1225 rightwards double arrow space space space space space space straight x squared equals 1225 rightwards double arrow space space space space space space straight x space equals space square root of 1225 rightwards double arrow space space space space space space straight x space equals space 35
    Hence the required values of x is 35.

    Question 259
    CBSEENMA10007925
    Wired Faculty App

    A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1 fourthm and a tread of 1 half m. (see the Fig.). Calculate the total volume of concrete required to build the terrace.
    Solution
    Volume of the concrete required to build

    Volume of concrete required to build the third

    Tips: -

    [Hint : Volume of concrete required to build the first step = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>]

    Question 260
    CBSEENMA10007927
    Wired Faculty App

    Which of the following list of numbers does form an A .P. ? If they form an A .P., write the next two terms.
    (i) 6, 1, –4, –9, –14............
    (ii) 1, –3, –5...........
    (iii) –2. 2. –2, 2, –2.............
    (iv) 1, 1, 1, 2, 2, 3, 3,.............
    Solution

    (i) a2 – a1 = 1 – 6 = –5
    a3 – a2 = –4 – 1 = –5
    a4 – a3 = –9 – (–4) = –9 + 4 = –5
    And, a5 – a3 = –14 – (–9) = –14 + 9 = 5
    Since, the difference between each pair of consecutive terms are constant.
    So, the given numbers form an A.P. with the common difference -5.
    The next two terms are :
    –14 + (–5) = –14 – 5 = –19
    –19 + (–5) = –24.
    (ii) a2 – a1 = –1 – 1 = –2
    a3 – a2 = –3 – (–1) = –3 + 1 = –2
    a4 – a3 = –5 – (–3) = –5 + 3 = –2
    Since, the difference between each pair of consecutive terms are constant.
    So, the given numbers form an A .P. with the common difference -2.
    The next two terms are :
    –5 + (–2) = –5 – 2 = –7
    –7 + (–2) = –7 – 2 = –9.
    (iii) a2 – a1 = 2 – (–2) = 2 + 2 = 4
    a3 –a2 = –2 – (2) = –2 – 2 = –4
    a4 – a3 = –5 – (–3) = –5 + 3 = –2
    Since, the difference between each pair of consecutive terms are not constant.
    So, the given numbers does not form an A .P.
    (iv) a2 – a, = 1 – 1 = 0
    a3 – a2 = 1 – 1 = 0
    a4 – a3 = 2 – 1 = 1
    a5 = 2 – 2 = 0
    a6 – a5 = 3 – 2 = 1
    a7 – a6 = 3 – 3 = 0
    Since, the difference between each pair of consecutive term are not constant.
    So, the given numbers does not form an A .P

    Question 261
    CBSEENMA10007928
    Wired Faculty App

    In which of the following situations, does the list of numbers involved make an arithmetic progression and why ?

    (i) The height (in cm) of some students of a school standing in a queue in morning assembly are 147, 148, 149........,157.

    (ii) The cash prizes (in Rs.) given by a institute to the papers of class X and XII are respectively, 500, 700, 900, ........ 1700.

    (iii) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.

    (iv) The cost of painting the wall of a building, when it costs Rs. 50 for the first metre and rises by Rs. 20 for subsequent metre.

    (v) The individual academic fee of the institute when it charge Rs. 500 for the first month and Rs. 100 for each subsequent month.
    Solution

    (i) Here, 147, 148, 149,....... 157
    a2 – a1 = 148 – 147 = 1
    a3 – a2 = 149 – 148 = 1
    Since, the difference between each pair of consecutive terms are constant.
    So, the given numbers form an A .P.
    (ii) Here, 500, 700, 900,.......... 1700
    a2 – ax = 700 – 500 = 200
    a3 – a2 = 900 – 700 = 200
    Since, the difference between each pair of consecutive terms are constant.
    So, the given numbers form an A .P.
    (iii) Here,
    Taxi fare for first km = 15
    for second km = 15 + 8=23
    for third km = 23 + 8 = 31
    So, we have 15, 23, 31,...........
    Since, the difference between each pair of consecutive terms are 8.
    So, the given numbers form an A .P.
    (iv) Cost of painting the wall of building for
    First metre = 50
    Cost of painting the wall of building for
    2nd metre = 50 + 20 = 70
    Cost of painting the wall of building for
    3rd metre = 70 + 20 = 90
    So, we have 50, 70, 90,...........
    Since, the difference between each pair of consecutive terms are 20.
    So, the given numbers form an A .P.
    (v) The academic fees for
    Ist month = 500
    Fees for 2nd month = 500 + 100 = 600
    Fees for 3rd month = 600 + 100 = 700
    So, we have 500, 600, 700,...........
    Since, the difference between each pair of consecutive terms are 100.
    So, the given numbers form an A .P.

    Question 262
    CBSEENMA10007929
    Wired Faculty App

    Write first four terms of the A .P., when the first term a and the common difference d are given as follows :

    (i) a = 10, d = 10.
    (ii) a = –1, d = 1/2
    (iii) a = –1.25, d = 0.25.
    (iv) a = p, d = –3q. 
    (v) a = b, d = 2c. 
    Solution

    Here, (i) a = 10, d = 10 then,
    First term = 10
    Second term = 10 + 10 = 20
    Third term = 20 + 10 = 30
    and fourth term = 30 + 10 = 40
    So, first four terms are 10, 20, 30, 40,...........

    Question 263
    CBSEENMA10007932
    Wired Faculty App

    Find the next term of the A. P.
    Which term of the A.P. 21, 18, 15.....is zero?
    Solution

    Here,   a = 21,  d = 18 - 21 = -3 and an = 0
    space space because      an = a + (n -1) d
    rightwards double arrow       0  = 21 + (n -1) (- 3)
    rightwards double arrow       3 (n - 1) = 21
    rightwards double arrow        3n = 24
    rightwards double arrow space space space space space straight n space equals space 24 over 3 equals 8
    Hence, eighth term is zero.

    Question 264
    CBSEENMA10007933
    Wired Faculty App

    Which term of the A .P. 14, 11, 8..... is - 1? 
    Solution

    Let nth term of the A.P. 14, 11, 8,.... is – 1.
    Here, a = 14, d = 11 – 14 = 8 – 11 = – 3 and an = – 1
    because space space space space space space space straight a subscript straight n equals straight a plus left parenthesis straight n minus 1 right parenthesis straight d rightwards double arrow space space space space space space space minus 1 space equals space 14 space plus space left parenthesis straight n minus 1 right parenthesis left parenthesis negative 3 right parenthesis rightwards double arrow space space space space space space space space minus 1 equals 14 minus 3 straight n plus 3 rightwards double arrow space space space space space space space 3 straight n minus 1 minus 17 equals 0 rightwards double arrow space space space space space space space space 3 straight n space equals space 18 rightwards double arrow space space space space space space space space straight n space equals 18 over 3 equals 6
    Hence, 6th term of the given A .P. is – 1.

    Question 265
    CBSEENMA10007934
    Wired Faculty App

     (i) For what value of p, are 2p – 1,7 and 3p, three consecutive terms of an A . P. ?

    (ii) For what value of p are 2p + 1, 13, 5p –3, three consecutive terms of an A . P. ? 
    Solution

    (i) 2p – 1, 7 and 3p will be three consecutive terms of an A .P. if
    Ilnd term – Ist term =IIIrd term – II term
    ⇒ 7 – (2p – 1) = 3p –7
    ⇒ 7 – 2p + 1 = 3p – 7
    ⇒ –2p – 3p = – 7 –8 ⇒ – 5p = – 15
    ⇒ p = 3.
    (ii) 2p + 1, 13 and 5p – 3 will be three consecutive terms of an AP if
    13 – (2p + 1) = (5p – 3) – 13
    ⇒ 13 – 2p – 1 = 5p – 3 – 13
    ⇒ –2p – 5p = –16 –12
    ⇒ –7p = –28
    ⇒ p = 4.

    Question 266
    CBSEENMA10007936
    Wired Faculty App

    Find the 10th term from the end of the following A .P. :
    (i) 1. 6, 11, 16............201
    (ii) 3, 6, 9, 12........... 102
    Solution

    (i) We have, l = 201,d = 5
    ∴ 10th term from the end
    = l – (n –1) d
    = 201 – (10 – 1)5
    = 201 – 45 = 156.
    (ii) We have, l = 102, d = 3,
    ∴ 10th term from the end
    = l – (n – 1)d
    = 102 – (10 – 1)3
    = 102 – 27 = 75.

    Question 267
    CBSEENMA10007942
    Wired Faculty App

    In an A .P., the First term is 8, nth term is 33 and sum to first n terms is 123. Find n and d, the common difference.

    Solution
    Here,   a1 = 8, an =33 and Sn = 123

    Hence, n = 6 and d = 5
    Question 268
    CBSEENMA10007943
    Wired Faculty App

    The first term of an A .P. is ‘p’ and its common difference is ‘q’ Find its 10thterm. 

    Solution

    Here, = p and d = q
    We know that an = a1 + (n – 1) d
    a10 = P + (10 – 1) q
    = p + 9qHence, 10th term of the A .P. is p + 9q. Ans.

    Question 270
    CBSEENMA10007946
    Wired Faculty App

    The nth term of an A .P. is 6n + 2 find its common difference.

    Solution

    We have, an = 6n + 2
    a1 = 6(1) + 2 = 6 + 2 = 8
    a2 = 6(2) + 2 = 12 + 2 = 14
    and, a3 = 6(3) + 2 = 18 + 2 = 20
    Now, common difference (d)
    = a3 – a2 = 20 – 14 = 6

    Question 271
    CBSEENMA10007947
    Wired Faculty App

    The nth term of an A .P. is 7 – 4n find its common difference.

    Solution

    We have, an = 7 – 4n
    an = 7 – 4 n
    an = 7 – 4(1) = 7 – 4 = 3
    a2 = 7 – 4 (2) = 7 – 8 = –1
    and, a3 = 7 – 4 (3) = 7 – 12 = –5
    Now, common difference (d) = a3 – a2 =– 5 + 1 = –4

    Question 272
    CBSEENMA10007948
    Wired Faculty App

    Write the common difference of an A .P. whose nth term is 3n + 5.

    Solution

    We have, an = 3n + 5
    a1 = 3(1) + 5 = 3 + 5 = 8
    a2 = 3 (2) + 5 = 6 + 5 = 11
    And, a3 = 3 (3) + 5 = 9 + 5 = 14
    Now, common difference = a2 – a1 = 11 – 8 = 3.

    Question 273
    CBSEENMA10007949
    Wired Faculty App

    For what value of k, are the numbers x, 2x + k and 3x + 6 three consecutive terms of an A .P.

    Solution

    x, 2x + k,3x + 6 are in A .P. if
    (2x + k) – x = (3x + 6) – (2x + k)
    ⇒ 2x + k – x = 3x + 6 – 2x – k
    ⇒ x + k = x + 6 – k
    ⇒ 2k = 6
    ⇒ k = 3 

    Question 274
    CBSEENMA10007950
    Wired Faculty App

    The 17th term of an A .P. exceeds its 10th term by 7. Find the common difference.

    Solution

    Let a be the first term and d be the common difference of the A .P.
    According to given statement,
    a17 = a10 +7
    ⇒ a + 16d = a + 9d + 7
    ⇒ 16d – 9d = 1
    ⇒ 7d = 7
    ⇒ d = 1.

    Question 275
    CBSEENMA10007951
    Wired Faculty App

    If 9th term of an A .P. is zero. Prove that 29th term is double of its 19th term. 

    Solution

     Let a be the first term and d be the common difference of the A .P. we have,
    a9 = 0 ⇒ a + 8d = 0 ⇒ a = – 8d
    Now, a29 = a + 28d = –8d + 28d = 20d
    2(a19) = 2(a + 18d) = 2(– 8d + 18d)
    = 2(10d) = 20d
    Hence, a29 = 2 (a19)

    Question 276
    CBSEENMA10007952
    Wired Faculty App

     If 5th term of an A .P. is zero. Prove that its 23rd term is three times its 11th term. 

    Solution

    Let a be the first term and d be the common difference of the A .P.
    We have,
    a5 = 0 ⇒ a + 4d = 0 ⇒ a = – 4d
    Now, a23 = a + 22d = –4d + 22d = 18d
    And 3 (a11) = 3(a + 10d)
    = 3(– 4d + 10d)
    = 3(6d) = 18d
    Hence, a23 = 3 (a11) Proved.

    Question 277
    CBSEENMA10007953
    Wired Faculty App

    How many numbers of two digits are divisible by 7?

    Solution

    Two digits numbers, which are divisible by
    7 are, 14, 21, 28.......... 98.
    Here, a = 14, d = 7 and l = 98
    We know that, Tn = a + (n –1)d
    ⇒ 98 = 14 + (n – 1) (7)
    [∵ l = Tn]
    ⇒ 98 = 14 + 7n – 7
    ⇒ 98 = 7n + 7
    ⇒ 98 = 7 = 7n
    ⇒ 91 = 7n ⇒ n = 13
    Hence, number of terms (n) = 13.

    Question 278
    CBSEENMA10007954
    Wired Faculty App

    Determine the A .P. whose 3rd term is 5 and the 7th term is 9.

    Solution

    We have, a3 = 5
    ⇒ a + (3 – 1) d = 5
    ⇒ a + 2d = 5 ...(i)
    and a7 = 9
    ⇒ a + (7– n) d = 9
    ⇒ a + 6d = 9 ...(ii)
    Subtracting (i) from (ii), we get
    (a + 6d) – (a + 2d) = 4
    ⇒ a + 6d – a – 2d = 4
    ⇒ 4d = 4
    ⇒ d = 1
    Putting the value of ‘d’ in (i), we get
    a + 2d = 5 ⇒ a + 2(1) = 5
    ⇒ a + 2 = 5 ⇒ a = 3
    Hence, the required A .P. be 3, 4, 5, 6, 7,...........

    Question 279
    CBSEENMA10007955
    Wired Faculty App

    Find the middle term of the A .P. 10,7,4, ...... (– 62).

    Solution

    We have, an = –62,
    a = 10 and d = –3 Now an = a + (n – 1)d
    ⇒ –62 = 10 + (n – 1) (– 3)
    ⇒ – 62 = 10 – 3n +3
    ⇒ – 62 = 13 – 3n
    ⇒ –3n = –75
    ⇒ n = 25
    Now middle term = a13 = a + 12d
    = 10 + 12(– 3)
    = 10 – 36 = – 26

    Question 280
    CBSEENMA10007957
    Wired Faculty App

     

    How many two digit numbers are divisible by 3.
    Solution

    We know, an = a + (n – 1)d
    ⇒ 99 = 12 + (n – 1) 3
    ⇒ 99 = 12 + 3n – 3
    ⇒ 99 = 9 + 3n
    ⇒ 3n = 90
    ⇒ n = 30Hence, there are 30 two-digit numbers which are divisible by 3.

    Question 281
    CBSEENMA10007966
    Wired Faculty App

     Write the common difference of an A .P. whose nth term is 3n + 5. 

    Solution

    We have, an = 3n + 5
    a1 = 3(1) + 5 = 3 + 5 = 8
    a2 = 3 (2) + 5 = 6 + 5 = 11
    And, a3 = 3 (3) + 5 = 9 + 5 = 14
    Now, common difference = a2 – a1 = 11 – 8 = 3.

    Question 282
    CBSEENMA10007967
    Wired Faculty App

    For what value of k, are the numbers x, 2x + k and 3x + 6 three consecutive terms of an A .P

    Solution

    x, 2x + k,3x + 6 are in A .P. if
    (2x + k) – x = (3x + 6) – (2x + k)
    ⇒ 2x + k – x = 3x + 6 – 2x – k
    ⇒ x + k = x + 6 – k
    ⇒ 2k = 6
    ⇒ k = 3

    Question 283
    CBSEENMA10007968
    Wired Faculty App

    The 17th term of an A .P. exceeds its 10th term by 7. Find the common difference.

    Solution

    Let a be the first term and d be the common difference of the A .P.
    According to given statement,
    a17 = a10 +7
    ⇒ a + 16d = a + 9d + 7
    ⇒ 16d – 9d = 1
    ⇒ 7d = 7
    ⇒ d = 1. 

    Question 284
    CBSEENMA10007969
    Wired Faculty App

    If 9th term of an A .P. is zero. Prove that 29th term is double of its 19th term. 

    Solution

     Let a be the first term and d be the common difference of the A .P. we have,
    a9 = 0 ⇒ a + 8d = 0 ⇒ a = – 8d
    Now, a29 = a + 28d = –8d + 28d = 20d
    2(a19) = 2(a + 18d) = 2(– 8d + 18d)
    = 2(10d) = 20d
    Hence, a29 = 2 (a19)

    Question 285
    CBSEENMA10007970
    Wired Faculty App

    If 5th term of an A .P. is zero. Prove that its 23rd term is three times its 11th term.

    Solution

    Let a be the first term and d be the common difference of the A .P.
    We have,
    a5 = 0 ⇒ a + 4d = 0 ⇒ a = – 4d
    Now, a23 = a + 22d = –4d + 22d = 18d
    And 3 (a11) = 3(a + 10d)
    = 3(– 4d + 10d)
    = 3(6d) = 18d
    Hence, a23 = 3 (a11) Proved.

    Question 286
    CBSEENMA10007971
    Wired Faculty App

    How many numbers of two digits are divisible by 7?

    Solution

    Two digits numbers, which are divisible by
    7 are, 14, 21, 28.......... 98.
    Here, a = 14, d = 7 and l = 98
    We know that, Tn = a + (n –1)d
    ⇒ 98 = 14 + (n – 1) (7)
    [∵ l = Tn]
    ⇒ 98 = 14 + 7n – 7
    ⇒ 98 = 7n + 7
    ⇒ 98 = 7 = 7n
    ⇒ 91 = 7n ⇒ n = 13
    Hence, number of terms (n) = 13.

    Question 287
    CBSEENMA10007974
    Wired Faculty App

    Which term of the A .P. 3, 15, 27, 39 ... will be 120 more than its 21st term?

    Solution
    Here,      a = 3,  d = 15 - 3 = 12

    Hence, 31st term is 120 more than its 21st term.
    Question 288
    CBSEENMA10007977
    Wired Faculty App

    Which term of the A .P. 4, 12, 20, 28, ... will be 120 more than its 21st term?

    Solution

    Here a = 4, d = 12 – 4 = 8. Then = a + 20d = 4 + 20 x 8 = 4 + 160 = 164
    Let an be 120 more than its 21st term
    therefore space space space space space space space straight a subscript straight n space equals space 164 plus 120 equals 284 rightwards double arrow space space space space space space space straight a left parenthesis straight n minus 1 right parenthesis straight d equals 284 rightwards double arrow space space space space space space 4 plus left parenthesis straight n minus 1 right parenthesis cross times 8 equals 284 rightwards double arrow space space space space space space 8 left parenthesis straight n minus 1 right parenthesis space equals space 280 rightwards double arrow space space space space space space space straight n minus 1 equals 280 over 8 equals 35 rightwards double arrow space space space space space space space space space space straight n space equals space 35 space plus space 1 space equals space 36
    Hence, 36th term is 120 more than its 21st term.

    Question 289
    CBSEENMA10007978
    Wired Faculty App

    Determine the A .P. whose 3rd term is 5 and the 7th term is 9.

    Solution

    We have, a3 = 5
    ⇒ a + (3 – 1) d = 5
    ⇒ a + 2d = 5 ...(i)
    and a7 = 9
    ⇒ a + (7– n) d = 9
    ⇒ a + 6d = 9 ...(ii)
    Subtracting (i) from (ii), we get
    (a + 6d) – (a + 2d) = 4
    ⇒ a + 6d – a – 2d = 4
    ⇒ 4d = 4
    ⇒ d = 1
    Putting the value of ‘d’ in (i), we get
    a + 2d = 5 ⇒ a + 2(1) = 5
    ⇒ a + 2 = 5 ⇒ a = 3
    Hence, the required A .P. be 3, 4, 5, 6, 7,..........

    Question 290
    CBSEENMA10007979
    Wired Faculty App

    How many two digit numbers which are multiples of 5 lie between 10 and 100 ?

    Solution

    The list of two digit numbers which are multiples of 5 are
    15, 20, 25, 30,...........95
    We have, a = 15, d = 5, an = 95
    We know, an = a + (n – 1) d
    ⇒ 95 = 15 + (n – 1)5
    ⇒ 95 = 15 + 5n – 5
    ⇒ 95 = 10 + 5n
    ⇒ 5n = 85
    ⇒ n = 17
    Hence, there are 17 two digit numbers lie between 10 and 100 which are multiples of 5.

    Question 291
    CBSEENMA10007980
    Wired Faculty App

    A sum of Rs. 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these, interests form an A .P. ? If so, find the interest at the end of 30 yrs.

    Solution

    Here,  P = 1000,  r = 8%
    Interest at the end of first year.
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    Interest at the end of second year
          equals fraction numerator straight P cross times straight r cross times straight t over denominator 100 end fraction equals space fraction numerator 1000 cross times 8 cross times 2 over denominator 100 end fraction equals R s. space 160
    Interest at the end of third year
    equals fraction numerator straight P cross times straight r cross times straight t over denominator 100 end fraction equals fraction numerator 1000 cross times 8 cross times 3 over denominator 100 end fraction equals space Rs. space 240
    So, we have 80, 160, 240, ..............
    Here, a = 80, d = 80
    Therefore, interest at the end of 30 yrs. i.e. a30 is,
    a30 = a + (30 –1) d = 80 + 29 x 80
    = 80 + 2320 = 2400
    So the interest at the end of 30 years is Rs. 2400.

    Question 292
    CBSEENMA10007981
    Wired Faculty App

    In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third and so on. There are 5 rose plant in the last row. How many rows are there in the flower bed?

    Solution

    The number of rose plants in the rows are 23, 21, 19, .......... 5
    Here, a = 23, d = –2, an = 5
    We know that,an = a + (n – 1)d
    ⇒ 5 = 23 + (n – 1) (–2)
    ⇒ 5 = 23 – 2n + 2
    ⇒ 5 = 25 – 2n
    ⇒ 2n = –20
    ⇒ n = 10
    Hence, there are 10 rows in the flower bed.

    Question 293
    CBSEENMA10007982
    Wired Faculty App

    If m times the mth term of an A .P. is equal to n times its nth term show that (m + n)th term of the A .P. is zero. 

    Solution

    Let ‘a’ be the 1st term and ‘d’ be the common difference of the A .P.
    mam = nan
    ⇒ m[a + (m – 1) d] = n[a + (n – 1) d]
    ⇒ m[a + md – d) = n[a + nd – d]
    ⇒ ma + m2d – md = na + n2d – nd
    ⇒ ma – na + m2d – n2d – md + nd = 0
    ⇒ (m – n)a + (m2 – n2)d – d(m – n) =0
    ⇒ (m – n)a + (m + n) (m – n)d – (m – n)d = 0
    ⇒ (m – n) [a + (m + n)d – d] = 0
    ⇒ a + (m + n)d – d = 0 ⇒ a + (m + n – 1)d = 0
    ⇒ am + n = 0
    Thus, the (m + n)th term of the given A .P. is zero.

    Question 294
    CBSEENMA10007991
    Wired Faculty App

    If the mth term of an A .P. is 1/n and the nth term is 1/m, find the (mn)th term.

    Solution
    Let ‘a’ be the first term and ‘d’ the common difference of the A .P.

    Question 295
    CBSEENMA10007993
    Wired Faculty App

    Split 69 into three parts such that they are in A .P. and the product of two smaller parts is 483.

    Solution

    Let the three parts, which are in A .P. be a – d, a, a + d
    Case I.
    a – d + a + a + d = 69
    ⇒ 3a = 69
    a = 23 ...(i)
    Case II. a (a – d) = 483 ...(ii)
    Putting the value of (i) in (ii), we get
    23 (23 - d)  = 483
    rightwards double arrow space space space space space 23 minus straight d space equals 483 over 23 rightwards double arrow space space space space space 23 straight d minus straight d space equals space 21 rightwards double arrow space space space space space space space space space minus straight d space equals space minus 2 rightwards double arrow space space space space space space space space space space space space straight d space equals space 2
    Thus, the three parts are
      a - d = 23 - 2 = 21 
          a  = 23
    and     a + d = 23 + 2 = 25

    Question 296
    CBSEENMA10007999
    Wired Faculty App

    The ratio of the 7th to the 3rd term of an A .P. is 12 : 5. Find the ratio of the 13th to 4th term.

    Solution
    Let ‘a’ be the first term and the common difference of the A .P.
    Question 297
    CBSEENMA10008000
    Wired Faculty App

    If the Pth term of an A .P. is q and qth term is P, prove that its nth term is (P + q – n).

    Solution

    Let ‘a’ be the first term and ‘d’ the common difference of the given A .P. Then,
    aP = q
    ⇒ a + (p – 1)d = q ...(i)
    a = P
    ⇒ a + (q – 1 )d = P ...(ii)
    Subtracting (ii) from (i), we get
    (a – a) + (p – 1)d – (q – 1)d = q – p
    ⇒ d(p – 1 – q + 1) = q – p
    d (p – q) = –(p – q)
    d = –1
    Putting the value of in (i), we get
    a + (p – 1)d = q
    ⇒ a + (p – 1) (–1) = q
    ⇒ a – p + 1 = q
    ⇒ a = p + q – 1
    Now, an = a + (n – 1)d
    = (p + 1) + (n – 1) (1)
    = p + q – 1 – n + 1 = p + q – n Proved.

    Question 298
    CBSEENMA10008001
    Wired Faculty App

    Find the 10th term from the end of the A .P. 8, 10, 12................ 126

    Solution

    Here, we have
    a = 8, d = 2, n = 2 and l = 126
    We know that,
    nth term from the end
    = l – (n – 1)d
    = 126 – (10 –1)2
    = 126 – 18
    = 108

    Question 299
    CBSEENMA10008003
    Wired Faculty App

    The sum of the 4th and 8th terms of an A .P. is 24 and the sum of the sixth and tenth terms is 44. Find the First three terms of the A .P.

    Solution

    It is given that,
    a4 + a8 = 24
    ⇒ a + 3d + a + 7d = 24
    ⇒ 2 a+ 10d = 24
    ⇒ a + 5d = 24 (i)
    And,
    a6 + a10 = 44
    ⇒ a + 5d + a + 9d = 44
    ⇒ 2a + 14d = 44
    ⇒ a + 7d = 22 (ii)
    solving (i) and (ii) we get
    a = –13 and d = 5
    Hence, the first three terms are
    a, a + d, a + 2d
    ⇒ – 13, (–13 + 5), (–13 + 2 x5)
    = –13, –8,–3
    ⇒ 2a + 14d = 44
    Problems Based on sum of ‘n’ th term of an A .P

    Question 300
    CBSEENMA10008007
    Wired Faculty App

    The first and the last term of an A .P. are 4 and 81 respectively. If the common difference is 7, how many terms are there in the A .P. and what is their sum?
    Solution
    Let ‘a’ be the first term, ‘d’ be the common difference and ‘l’ be the last term. Then.

    Hence, there are 12 terms in the A .P. and their sum is 510.
    Question 301
    CBSEENMA10008009
    Wired Faculty App

    If the sum of first 7 terms of an A .P. is 49 and that of first 17 terms is 289, find the sum of first n terms.

    Solution
    Let ‘a’ be the first term and ‘d’ be the common difference of the given A .P. Then,
    S7 = 49 and S17 = 289

    Question 302
    CBSEENMA10008011
    Wired Faculty App

    If Sn, the sum of first n terms of an A .P. is given by Sn = 3n2 – 4n, then find its nth term.

    Solution

    Sn = 3n2 –4n
    ⇒ Sn – 1= 3 (n – 1)2 –4 (n – 1)
    ∴ Sn – Sn –1 = (3n2 – 4n) – {3(n – 1)2 – 4 (n –1)}
    ⇒ an = (3n2 – 4n) – {3(n2 – 2n+ 1) – 4n ± 4} = 3n2 – 4n – {3n2 – 6n + 3 – 4n + 4}
    = 3n2 – 4n – {3n2 – 10n + 7}
    = 3n2 – 4n – 3n2 + 10n – 7
    = 6n – 7
    Hence, required nth term is 6n – 7. 

    Question 303
    CBSEENMA10008013
    Wired Faculty App

    If Sn, the sum of first n terms of an A .P. is given by Sn = 5n2 + 3n, then find its nthterm.
    Sn = 5n2 + 3n
    Solution

    ⇒ Sn – 1 = 5(n – 1)2 + 3(n – 1)
    ∴ Sn – Sn – 1 = (5n2 + 3n) – [5(n – 1)2 + 3(n – 1)}
    ⇒ an = (5n2 + 3n) – {5 (n2 – 2n + 1) + 3n – 3}
    = (5n2 + 3n) – {5n2 – 10n + 5 + 3n –3}
    = 5n2 + 3n – 5n2 + 7n – 2
    = 10n – 2
    Hence, required nth term is 10n – 2.

    Question 304
    CBSEENMA10008017
    Wired Faculty App

    Find the sum of the :
    First 25 terms of an A .P. whose nth term is given by an= 7 – 3n. z

    Solution

    Sn = 5n2 + 3n
    ⇒ Sn – 1 = 5(n – 1)2 + 3(n – 1)
    ∴ Sn – Sn – 1 = (5n2 + 3n) – [5(n – 1)2 + 3(n – 1)}
    ⇒ an = (5n2 + 3n) – {5 (n2 – 2n + 1) + 3n – 3}
    = (5n2 + 3n) – {5n2 – 10n + 5 + 3n –3}
    = 5n2 + 3n – 5n2 + 7n – 2
    = 10n – 2
    Hence, required nth term is 10n – 2

    Question 305
    CBSEENMA10008019
    Wired Faculty App

    Find the sum of the :
    (i) First 25 terms of an A .P. whose nth term is given by an= 7 – 3n
    Solution

    We have that,
    an = 7 – 3n
    If n = 1, a1 = 7 – 3n
    = 7 – 3 = 4
    n = 2, a2 = 7 – 3(2)
    = 7 – 6 = 1
    n = 3, = 7 – 3(3)
    = 7 – 9 = –2
    n = 4, a4 = 7 – 3(4)
    = 7 – 12 = –5
    So, we have following A .P.’s 4, 1, – 2, – 5, ..........
    Here, a = 4, d = –3
    We space know comma space straight S subscript straight n space equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket So comma space space straight S subscript 25 equals 25 over 2 left square bracket negative 2 minus 72 right square bracket equals 25 over 2 cross times negative 74 equals 25 cross times negative 37 equals negative 925

    Hence, the sum of first 25 terms of an A .P.
    be –925.

    Question 306
    CBSEENMA10008022
    Wired Faculty App

    The sum of n terms of 2 A .P.’s sire in the ratio 7n + 1 : 4n + 27, show that the ratio of their 11th term is 4 : 3.

    Solution
    Let the first and second term of the A .P. be a1 and a2 and common difference be d1and d2 It is given that :

    Question 307
    CBSEENMA10008025
    Wired Faculty App

    If the sum of m terms of an A .P. is the same as the sum of its nth terms. Show that the sum of its (m + n)th terms is zero.

    Solution
    Let ‘a’ be the first term and ‘d’ the common difference of the A .P.

    Question 308
    CBSEENMA10008028
    Wired Faculty App

    If S1, S2,S3, be the sum of n, 2n and 3n terms S3 = 3(S1 – S1).

    Solution
    Let ‘a’ and ‘d’ be the first term and common difference of the A .P.

    Hence L.H.S. = R.H.S.
    Question 309
    CBSEENMA10008032
    Wired Faculty App

    The sum of the first p, q and r terms of an A .P. are a, b, c respectively.
    Prove that  straight a over straight p left parenthesis straight q minus straight r right parenthesis plus straight b over straight q left parenthesis straight r minus straight p right parenthesis plus straight c over straight r left parenthesis straight p minus straight q right parenthesis equals 0
    Solution
    Let ‘a’ be the first term and d the common difference of the A .P.
    Question 310
    CBSEENMA10008034
    Wired Faculty App

    If in sin A .P. the sum of m terms is equal to n and the sum of n terms is equal to m, then prove that the sum of (m + n) terms is –(m + n).
    Solution
    Let ‘a’ be the first term and ‘d’ the common difference of the A .P.
    Question 311
    CBSEENMA10008044
    Wired Faculty App

    The ratio of the sums of and ‘n’ terms of an A .P. is m2 : n2. Show that the ratio of the mth and nth terms is (2m – 1) : (2n – 1).

    Solution
    Let ‘a’ be the first term and ‘d’ be the common difference of the A .P.

    Question 313
    CBSEENMA10008067
    Wired Faculty App

    Find the sum of the series :

    4 + 8 + 12 +......... + 100.

    Solution
    Let ‘b’ be the first term and ‘d’ the common difference of the A .P.
    a = 4, d = 4,  an = 100 =  1
    We know that,  an = a+ (n -1)d
    rightwards double arrow   100 = 4 + (n - 1) d
    rightwards double arrow   100 = 4 + (n - 1) 4
    rightwards double arrow    4n = 100
    rightwards double arrow    n = 25
    And comma space space straight S subscript straight n equals straight n over 2 left square bracket straight a plus straight l right square bracket space space space left square bracket because space straight l space equals straight T subscript straight n right parenthesis right square bracket So comma space space space space straight S subscript 25 equals 25 over 2 space space space left square bracket 4 space plus space 100 right square bracket space space space space space space space space space space space space space space equals 25 over 2 cross times 104 equals 25 cross times 52 space space space space space space space space space space space space space space space equals space 1300
    Question 314
    CBSEENMA10008071
    Wired Faculty App

    Find the sum of the series :

    7 + 10 1/2 + 14 + .. .. + 84.

    Solution
    Let ‘a’ be first term and ‘d’ be the common difference then,

    Question 315
    CBSEENMA10008074
    Wired Faculty App

    Find the sum of the series :
    34 + 32 + 30 +........ + 10

    Solution
    Let ‘a’ be the firs term and ‘d’ the common difference, then
    a = 34,    d = - 2, an = 10 = l
    We know that,  an = a + (n - 1) d
    rightwards double arrow   10 = 34 + (n - 1) (- 2)
    rightwards double arrow   10 = 34 - 2n + 2
    rightwards double arrow   10 = 36 - 2n
    rightwards double arrow   - 26 = - 2n
    rightwards double arrow      n = 13
    And,   S subscript n equals n over 2 left square bracket a space plus space l k right square bracket S o comma space S subscript 13 equals 13 over 2 left square bracket 34 plus 10 right square bracket equals 13 over 2 cross times 44 equals 13 cross times 22 equals 286
    Question 316
    CBSEENMA10008078
    Wired Faculty App

    Find the sum of the series :
    –5 + (–8) + (–11) +........ + (–230).

    Solution
    Let ‘a’ be the first term and d the common difference then,
    a = - 5,  d = - 3, an = - 230 = l
    We know that, an = a + (n - 1) d
    rightwards double arrow -230 = - 5 + (n - 1) ( - 3)
    rightwards double arrow - 230 = -5 - 3n + 3
    rightwards double arrow - 230 = -2 - 3n
    rightwards double arrow - 228 = -3n
    rightwards double arrow space space straight n space equals space 228 over 3 equals 76
    We space know space that comma space straight S subscript straight n equals straight n over 2 left square bracket straight a plus straight l right square bracket So comma space space space straight S subscript 16 equals 76 over 2 left square bracket negative 5 space plus space left parenthesis negative 230 right parenthesis right square bracket space space space space space space space space space space space space space equals space 38 space left parenthesis negative 235 right parenthesis space minus 8930.
    Question 317
    CBSEENMA10008080
    Wired Faculty App

    Find the sum of all natural numbers between 100 and 1000 which arc multiples of 7.

    Solution

    The First number between 100 and 1000 which is multiple of 7 be 105 and last number 994.
    Thus we have following A .P.’s

    Question 318
    CBSEENMA10008081
    Wired Faculty App

    Tanvika purchases every year National Saving Certificates of value exceeding the last year purchases by Rs. 25. After 20 years, he find the total value of certificates purchased by him is Rs. 7250. Find the value of the certificates purchased.

    (i) in the first year. (ii) in the 12th year.
    Solution
    Let the value of certificates purchased in the first year be Rs. ‘a’ and annual increase = Rs.25. Here, first term = a, d = 25, S20 = 7250.
    We know that,
     straight S subscript straight n equals straight n over 2 left square bracket 2 straight a space plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket rightwards double arrow space straight S subscript 10 equals 20 over 2 left square bracket 2 straight a plus left parenthesis 20 minus 1 right parenthesis 25 right square bracket rightwards double arrow space 7250 space equals space 10 space left parenthesis 2 straight a plus 19 cross times 25 right parenthesis

    ⇒ 7250 = 10 (2a + 475)
    ⇒ 725 = 2a + 475 ⇒ 2a = 250
    ⇒ a = 125
    Hence, the value of certificates purchased in first year = Rs. 125.
    Now, value of certificates purchased in 12th year.
    a12 = a + (12 – 1 )d = 125 + (11) (25) = 125 + 275 = Rs. 400.
    Question 320
    CBSEENMA10008083
    Wired Faculty App

    Nimit saves Rs. 50 during a month, Rs. 100 during next month, Rs. 150 in next month, if he continues his savings in this way, in how many months will he save Rs. 10500.

    Solution

    Sequence of savings is 50, 100, 150,.........
    Since, common difference between any two consecutive terms is constant (i.e., 50).
    So, given sequence are in A .P.
    Here, a = 50, d = 50, Sn = 10,500

    Since, n = –21 is not possible.
    Hence an = 20
    i.e., In 20 months he will save Rs. 10500.

    Question 321
    CBSEENMA10008084
    Wired Faculty App

    A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows : Rs. 200 for 1st day, Rs. 250 for second day, Rs. 300 for third day and so on. If the contractor pays Rs. 27750 as penalty, find the number of days for which the construction work is delayed. 

    Solution

    Given penalties are, Rs. 200, Rs. 250, Rs. 300 ......
    Clearly, it is an arithmetic progression with first term a = 200, common difference d = 50 and Sn = Rs. 27750.

    Since, n = –37 is rejected. Hence, the construction work is delayed for 30 days.

    Question 322
    CBSEENMA10008085
    Wired Faculty App

    From your pocket money, you save Rs. 1 on day 1, Rs. 2 on day 2, Rs. 3 on day 3 and so on. How much money will you save in (he month of March 2008?

    Solution

    Savings in Rs. be 1, 2, 3, 4,................ The sequence 1, 2, 3, 4,............is an A. P. with first term (a) = 1 and common difference (d) 1.
    Here, total number of days in March = 31 i.e., n = 31
    Thus, we have a = 1, d = 1 and n = 31.
    Now comma space space space straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket rightwards double arrow space space space space space straight S subscript 31 equals 31 over 2 left square bracket 2 space straight x space 1 space plus space left parenthesis 31 minus 1 right parenthesis 1 right square bracket space space space space space space space space space space space space space space space equals space 31 over 2 left square bracket 2 plus 30 right square bracket space space space space space space space space space space space space space space space space equals 31 over 2 cross times 31 equals 31 cross times 16 equals 496
    Hence, amount of savings in March 2008 in 496.

    Question 323
    CBSEENMA10008086
    Wired Faculty App

    If the 10th term of an A .P. is 47 and its first term is 2, find the sum of its first 15 terms. 

    Solution
    Let a be the first term and d be the common difference of the given A .P.

    Question 326
    CBSEENMA10008094
    Wired Faculty App

    Find the common difference of the A .P. and write the next two terms :

    Solution

    Solution not provided.
    Ans.

    Question 330
    CBSEENMA10008101
    Wired Faculty App

    Write first four terms of the A .P, when the first term a and the common difference d are given as follows :

    a = p,  d = 50
    Solution

    Solution not provided.
    Ans,  p, p + 50, p + 100, p + 150

    Question 332
    CBSEENMA10008103
    Wired Faculty App

    Write first four terms of the A .P, when the first term a and the common difference d are given as follows :

    a = 10.0,  d = 0.5
    Solution

    Solution not provided.
    Ans,  10.0, 10.5, 11, 11.5

    Question 333
    CBSEENMA10008105
    Wired Faculty App

    For the following A .Ps, write the first term and the common difference :
    2, 5, 8, 11, ................
    Solution

    Solution not provided.
    Ans. a = 2, d = 3

    Question 334
    CBSEENMA10008106
    Wired Faculty App

    For the following A .Ps, write the first term and the common difference :
    1, 6, 11, 16, ...........
    Solution

    Solution not provided.
    Ans. a = 1, d = 5

    Question 335
    CBSEENMA10008107
    Wired Faculty App

    For the following A .Ps, write the first term and the common difference :
    -17, -10, -3, 4, .........
    Solution

    Solution not provided.
    Ans. a = 17, d = 17

    Question 336
    CBSEENMA10008108
    Wired Faculty App

    For the following A .Ps, write the first term and the common difference :
    0, -4, -8, ........
    Solution

    Solution not provided.
    Ans. a = 0, d = 4

    Question 337
    CBSEENMA10008109
    Wired Faculty App

    For the following A .Ps, write the first term and the common difference :
    4, 0, -4
    Solution

    Solution not provided.
    Ans. a = 4, d = - 4

    Question 338
    CBSEENMA10008110
    Wired Faculty App

    For the following A .Ps, write the first term and the common difference :
    -14, -19, -24, ........
    Solution

    Solution not provided.
    Ans. a = - 14, d = - 5

    Question 339
    CBSEENMA10008112
    Wired Faculty App

    For the following A .Ps, write the first term and the common difference :
    1, -1, -3, -5, .........
    Solution

    Solution not provided.
    Ans. a = 1, d = 2

    Question 340
    CBSEENMA10008121
    Wired Faculty App

    Write first four terms the A .P. when the first term ‘a’ and the common difference ‘d’ are given as follows :

     a = 4, d = – 3

    Solution

    a = 4, d = –3
    First term = a = 4
    Second term = 4 + d = 4 + (–3) = 1
    Third term = 1 + d = 1 + (–3) = –2
    Fourth term = –2 + d = –2 + (–3) = –5
    Hence, four first terms of the given A .P. are 4, 1, -2, -5.

    Question 341
    CBSEENMA10008122
    Wired Faculty App

    The 17th term of an A .P. exceeds the 10th term by 7. Find the common difference.

    Solution

    We have
    a17 = a + 16d
    and a10 = a + 9d
    According to the given question,
    a17 – a10 = 7
    ⇒ (a + 16d) – (a + 9d) = 7
    ⇒ a + 16d – a – 9d = 7
    ⇒ 7d – 7
    ⇒ d = 1
    Hence, common difference is 1

    Question 342
    CBSEENMA10008123
    Wired Faculty App

    Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 100th terms?

    Solution

    Let the common difference of both the
    A.P.’s = d
    and First term of Ist A.P. = a
    also first term of 2nd A .P. = b
    Then, 200th term of 1st A .P. = a + 99d
    and 100th term of IInd A .P. = b + 99d
    Difference between their 100th term = 100
    (given)
    ⇒ (a + 99d) – (b + 99d) = 100
    ⇒ a – b = 100 ...(i)
    Now, a1000 of Ist A.P. = a + 999d
    and a1000 of Ilnd A .P. = b + 999d
    Hence, difference between their 1000th terms
    = (a + 999d) (b + 999d) = a – b
    = 100. [Using (i)]

    Question 343
    CBSEENMA10009584
    Wired Faculty App

    For what value of k will k+9, 2k-1 and 2k+7 are the consecutive terms of an A.P?
    Solution

    If k+9, 2k-1 and 2k+7 are the consecutive terms of A.P, then the common difference will be the same.
    ∴ (2k – 1) – (k + 9) = (2k + 7) – (2k – 1)

    ∴ k – 10 = 8

    ∴ k = 18

    Question 344
    CBSEENMA10009591
    Wired Faculty App

    The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
    Solution

    We have given that
    4th term of an A.P.= a4 = 0
    ∴ a + (4 – 1)d = 0
    ∴ a + 3d = 0
    ∴ a = –3d        ….(1)

    25th term of an A.P. = a25
    = a + (25 – 1)d
    = –3d + 24d      ….[From the equation (1)]
    = 21d

    3 times 11th term of an A.P. = 3a11
    = 3[a + (11 – 1)d]
    = 3[a + 10d]
    = 3[–3d + 10d]
    = 3 × 7d
    = 21d

    ∴ a25 = 3a11

    i.e., the 25th term of the A.P. is three times its 11th term.

    Question 345
    CBSEENMA10009597
    Wired Faculty App

    If the ratio of the sum of first n terms of two A.P’s is (7n +1): (4n + 27), find the ratio of their mth terms.
    Solution

    Let a1, a2 be the first terms and d1 , d2 the common differences of the two given A.P's.
    Then, we have Sn = straight S subscript straight n space equals straight n over 2 left square bracket 2 straight a subscript 1 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 1 right square bracket space and space straight S subscript straight n equals straight n over 2 left square bracket 2 straight a subscript 2 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 2 right square bracket therefore straight S subscript straight n over straight S subscript straight n equals fraction numerator straight n over 2 left square bracket 2 straight a subscript 1 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 1 right square bracket over denominator straight n over 2 left square bracket 2 straight a subscript 2 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 2 right square bracket end fraction space equals fraction numerator 2 straight a subscript 1 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 1 space over denominator 2 straight a subscript 2 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 2 end fraction It space is space given space that space straight S subscript straight n over straight S subscript straight n space equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction therefore space fraction numerator 2 straight a subscript 1 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 1 space over denominator 2 straight a subscript 2 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 2 end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction space.. space left parenthesis 1 right parenthesis To space find space the space ratio space of space the space straight m to the power of th space terms space of space the space tow space given space straight A. straight P apostrophe straight s space replace space straight n space by space left parenthesis 2 straight m minus 1 right parenthesis space in space equation space 1 therefore space equals fraction numerator 2 straight a subscript 1 plus left parenthesis 2 straight m minus 2 right parenthesis straight d subscript 1 over denominator 2 straight a subscript 2 plus left parenthesis 2 straight m minus 2 right parenthesis straight d subscript 2 end fraction equals fraction numerator 14 straight m minus 7 plus 1 over denominator 8 straight m minus 4 plus 27 end fraction therefore space fraction numerator straight a subscript 1 plus left parenthesis straight m minus 1 right parenthesis straight d subscript 1 over denominator straight a subscript 2 left parenthesis straight m minus 1 right parenthesis straight d subscript 2 end fraction space equals space fraction numerator 14 straight m minus 6 over denominator 8 straight m plus 23 end fraction Hence comma space the space ratio space of space the space straight m to the power of th space terms space of space the space two space straight A. straight P apostrophe straight s space is space 14 straight m minus 6 colon 8 straight m plus 23

    Question 346
    CBSEENMA10009609
    Wired Faculty App

    The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to the sum of the numbers of houses following X.
    Solution

    Let there be a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it.
    That is , 1+2+3+......+(x-1) =(x-1)+(x+2)+.....+49
    therefore,
    1+2+3+.....+(x-1)
    =[1+2+.....+x+(x+1)+.....+49]-(1+2+3+....+x)
    therefore space fraction numerator straight x minus 1 over denominator 2 end fraction left square bracket 1 plus straight x minus 1 right square bracket space equals 49 over 2 left square bracket 1 plus 49 right square bracket minus straight x over 2 left square bracket 1 plus 2 right square bracket therefore straight x left parenthesis straight x minus 1 right parenthesis equals 49 space straight x 50 minus straight x left parenthesis 1 plus straight x right parenthesis therefore straight x left parenthesis straight x minus 1 right parenthesis plus 1 left parenthesis 1 plus straight x right parenthesis equals 49 straight x 50 straight x squared minus straight x plus straight x plus straight x squared space equals space 49 space straight x 50 therefore space straight x squared space equals 49 space straight x 25 solving space saqure straight x space equals space 7 space straight x 5 space equals 35
    Since x is not a fraction, the value of x satisfying the given condition exists and is equal to 35.

    Question 347
    CBSEENMA10009621
    Wired Faculty App

    In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP,  Where Sn denotes the sum of its first n terms.
    Solution

    Let a and d be the first term and the common difference of the AP, respectively
    therefore space Sum space of space straight n space terms comma space straight S subscript straight n space equals space straight n over 2 space left square bracket 2 straight a space plus space left parenthesis straight n minus 1 right parenthesis straight d right square bracket We space have straight S subscript 5 space plus space straight S subscript 7 space equals space 167 rightwards double arrow space 5 over 2 space left parenthesis 2 straight a space plus space 4 straight d space right parenthesis space plus 7 over 2 space left parenthesis 2 straight a space plus space 6 straight d right parenthesis space equals space 167 rightwards double arrow space 5 space left parenthesis straight a space plus space 2 straight d right parenthesis space space plus 7 space left parenthesis straight a space plus 3 straight d right parenthesis space equals 167 rightwards double arrow 12 straight a space space plus space 31 straight d space equals space 167 space space.. left parenthesis straight i right parenthesis Also comma straight S subscript 10 space equals space 235 rightwards double arrow space 10 over 2 space left parenthesis 2 straight a space plus 9 straight d right parenthesis space equals space 23 rightwards double arrow 2 straight a space plus 9 straight d space equals space 47 space space... space left parenthesis ii right parenthesis solving space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space ge straight a equals space 1 space and space straight d equals 5 Hence comma space required space AP space is space 1 comma space 6 comma space 11 space space

    Question 348
    CBSEENMA10009624
    Wired Faculty App

    The 14th term of an AP is twice its 8th term. If its 6th  terms is -8, then find the sum of its first 20 terms.
    Solution

    Let a and d be the first term and the common difference of the AP, respectively.
    ∴ nth term of the AP, an = a + (n-1)d
    So,
    a14 = a+ (14-1)d = a + 13d
    a8 = a + (8 - 1)d = a + 7d
    a6 = a+ (6 - 1)d = a + 5d
    According to the question,
    a14 = 2a8
     ⇒ a + 13d = 2 (a + 7d)
    ⇒ a + d = 0 .... (i)
    Also,
    a6 = a + 5d = - 8 ... (ii)
    solving (i) and (ii), we get
    a = 2 and d = -2
    ∴ S20 = 20/2 [2 x 2 + (20 - 1)(-2)]
    = - 340
    Hence, the sum of the first 20 terms is -340.

    Question 349
    CBSEENMA10009635
    Wired Faculty App

    Find the 60th term of the AP 8, 10,12, ..., if it has a total of 60 terms and hence find the sum of its last 10 terms.
    Solution

    The given AP is 8, 10, 12, ....
    So,
    First term =a = 8
    Common difference = d = 10-8 =2
    We know that nth term of an AP, an = a + (n - 1)
    60th term of the given AP = a60 = 8 +( 60-1) x  2 = 8 + 59 x 2 = 8 + 118 = 126
    Therefore, the 60th term of the given AP is 126
    It is given that the AP has a total of 60 terms. So, in order to find sum of last n terms. we take
    First term, A = 126
    Common difference, D = -2
    Now,
    Sum space of space straight n space term space of space AP space from space the space end space equals space straight n over 2 left square bracket 2 straight A space plus left parenthesis straight n space minus space 1 right parenthesis space straight D right square bracket therefore space Sum space of space last space 10 space terms space of space the space given space AP equals space 10 over 2 left square bracket space 2 space straight x space 126 space plus space left parenthesis 10 space minus space 1 space right parenthesis space straight x space left parenthesis negative 2 right parenthesis right square bracket equals space 5 space left square bracket 252 space plus 9 space straight x space left parenthesis negative 2 right parenthesis right square bracket equals space 2 space left parenthesis 252 space minus 18 right parenthesis equals space 5 space straight x space 234 equals space 1170 hence space comma space the space sum space of space last space 10 space terms space of space the space given space AP space is space 1170

    Question 350
    CBSEENMA10009649
    Wired Faculty App

    Find how many integers between 200 and 500 are divisible by 8.
    Solution

    The first term between 200 and 500 divisible by 8 is 208, and the last term is 496.
    So, first term (a) = 208
    Common difference (d) = 8
    an=a+(n−1)d=496
    ⇒208+(n−1)8=496
    ⇒(n−1)8=288
    ⇒n−1=36⇒n=37
    Hence, there are 37 integers between 200 and 500 which are divisible by 8.

    Question 351
    CBSEENMA10009654
    Wired Faculty App

    If the mth term of an A. P. is 1 over straight n and nth term is 1 over straight m then show that its (mn)th term is 1. 
    Solution

    Let x and d be the first term and common difference respectively of the AP, respectively.
    Then,
    straight m to the power of th space term space equals space 1 over straight n rightwards double arrow space straight x plus left parenthesis straight n minus 1 right parenthesis straight d space equals space 1 over straight n straight x space equals 1 over straight n minus left parenthesis straight m minus 1 right parenthesis straight d space.... space left parenthesis 1 right parenthesis straight n to the power of th space term space equals 1 over straight m rightwards double arrow straight x space plus left parenthesis straight n minus 1 right parenthesis straight d space equals space 1 over straight m rightwards double arrow space 1 over straight n space minus space left parenthesis straight m minus 1 right parenthesis straight d space plus left parenthesis straight n minus 1 right parenthesis straight d space equals space 1 over straight m space left parenthesis from space equ space 1 right parenthesis rightwards double arrow straight d space left parenthesis straight n minus 1 minus straight m plus 1 right parenthesis space equals space 1 over straight m minus 1 over straight n rightwards double arrow space straight d left parenthesis straight n minus straight m right parenthesis space equals space fraction numerator straight n minus straight m over denominator nm end fraction space rightwards double arrow space straight d space equals 1 over mn Putting space value space of space straight d space in space equation space left parenthesis 1 right parenthesis comma space we space have straight x space equals space 1 over straight n space minus left parenthesis straight m minus 1 right parenthesis 1 over mn space equals space 1 over straight n minus straight m over mn space plus 1 over mn therefore space straight t subscript mn space equals straight a space plus space left parenthesis mn minus 1 right parenthesis straight d space equals space 1 over mn space plus space left parenthesis mn minus 1 right parenthesis 1 over mn space equals space 1

    Question 352
    CBSEENMA10009656
    Wired Faculty App

    Find the sum of n terms of the series  open parentheses 4 minus 1 over straight n close parentheses space plus open parentheses 4 minus 2 over straight n close parentheses space plus open parentheses 4 minus 3 over straight n close parentheses space plus.........
    Solution
    S space equals open parentheses 4 minus 1 over straight n close parentheses space plus open parentheses 4 minus 2 over straight n close parentheses space plus open parentheses 4 minus 3 over straight n close parentheses space plus......... space equals space left square bracket 4 space plus space 4 space plus space 4 space plus...... right square bracket space minus open square brackets 1 over straight n space plus 2 over straight n plus 3 over straight n plus.... close square brackets space equals space 4 left square bracket 1 plus 1 plus 1 plus.... right square bracket space minus 1 over straight n left square bracket 1 plus 2 plus 3 plus.... right square bracket space equals space straight S subscript 1 minus straight S subscript 2 straight S subscript 1 space equals space 4 space left square bracket 1 plus 1 plus 1 plus.... right square bracket straight a space equals 1 space comma straight d space equals 0 straight S subscript 1 space equals space 4 space straight x straight n over 2 left square bracket 2 space straight x 1 space plus left parenthesis straight n minus 1 right parenthesis space straight x space 0 right square bracket open parentheses straight S subscript straight n space equals space straight n over 2 left parenthesis 2 straight a space plus left parenthesis straight n minus 1 right parenthesis straight d right parenthesis close parentheses straight S subscript 1 space equals space 4 straight n straight S subscript 2 space equals space 1 over straight n left square bracket 1 plus 2 plus 3 plus...... right square bracket straight a space equals 1 comma space straight d equals 2 minus 1 space equals space 1 straight S subscript 2 space equals space 1 over straight n space straight x straight n over 2 left square bracket 2 space straight x 1 space plus left parenthesis straight n minus 1 right parenthesis space straight x 1 right square bracket space equals space 1 half left square bracket 2 plus straight n minus 1 right square bracket space equals space 1 half left square bracket 1 plus straight n right square bracket Thus comma space straight S space equals space straight S subscript 1 space minus straight S subscript 2 space equals space 4 straight n space minus 1 half space left square bracket 1 plus straight n right square bracket straight S space equals space fraction numerator 8 straight n minus 1 minus straight n over denominator 2 end fraction space equals space fraction numerator 7 straight n minus 1 over denominator 2 end fraction
    Question 353
    CBSEENMA10009667
    Wired Faculty App

    The ratio of the sums of the first m and first n terms of an A. P. is m2: n2. Show that the ratio of its mth and nth terms is (2m−1):(2n−1). 
    Solution

    Let Sm and Sn be the sum of the first m and first n terms of the A.P. respectively. Let, a be the first terms and d be a common difference.
    straight S subscript straight m over straight S subscript straight n space equals straight m squared over straight n squared rightwards double arrow space fraction numerator begin display style straight m over 2 open square brackets 2 straight a plus left parenthesis straight m minus 1 right parenthesis straight d close square brackets end style over denominator begin display style straight n over 2 open square brackets 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d close square brackets end style end fraction space equals space straight m squared over straight n squared rightwards double arrow space fraction numerator 2 straight a space plus left parenthesis straight m minus 1 right parenthesis straight d over denominator 2 straight a left parenthesis straight n minus 1 right parenthesis straight d end fraction space equals space straight m over straight n
    ⇒[2a+(m−1)d]
    n =[2a(n−1)d]m
    ⇒2an+mnd−nd=2am+mnd−md
    ⇒md−nd=2am−2an
    ⇒(m−n)d=2a(m−n)
    ⇒d=2a
    Now, the ratio of its mth and nth terms is
    straight a subscript straight m over straight a subscript straight n space equals space fraction numerator straight a space plus left parenthesis straight m minus 1 right parenthesis straight d over denominator straight a plus left parenthesis straight n minus 1 right parenthesis straight d end fraction space equals space space fraction numerator begin display style straight a space plus left parenthesis straight m minus 1 right parenthesis 2 straight a end style over denominator begin display style straight a plus left parenthesis straight n minus 1 right parenthesis 2 straight a end style end fraction space equals space fraction numerator straight a space left parenthesis 1 plus 2 straight m minus 2 right parenthesis over denominator straight a left parenthesis 1 plus 2 straight n minus 2 right parenthesis end fraction space equals space fraction numerator 2 straight m minus 1 over denominator 2 straight n minus 1 end fraction
    Thus, the ratio of its mth and nth terms is 2m – 1 : 2n – 1.

    Question 354
    CBSEENMA10009678
    Wired Faculty App

    In an AP, if the common difference (d) = –4, and the seventh term (a7) is 4, then find the first term.
    Solution

    a7 = 4

    a + 6d = 4 

    as an = a + (n-1) d

    but d = -4

    a + (-24) = 4
    a = 4 + 24 = 28
    therefore first term a = 28

    Question 355
    CBSEENMA10009687
    Wired Faculty App

    Find the sum of first 8 multiples of 3.
    Solution

    First 8 multiples of 3 are

    3, 6, 9, 12, 15, 18, 21, 24

    The above sequence is an A.P.

    a = 3, d = 3 and last term l = 24

    Sn = n2 (a + l)  = 82 [ 3 + 24]  = 4 (27)Sn = 108

    Question 356
    CBSEENMA10009710
    Wired Faculty App

    The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15. Find the numbers.
    Solution

    Let the numbers be (a -3d),(a-d), (a+d) and (a+3d)

    ∴ (a-3d) + (a-d) +(a+ d) + (a +3d) = 32

    ⇒ 4a = 32

    a = 8

    Also, (a-3d)(a+ 3d)(a-d)(a+d) =715⇒ 15a2 -135d2 = 7a2 -7d2⇒ 8a2 = 128d2d2 = 8a2128 = 8 x 8x8128d2 =4d =±2

    If d =2 numbers are : 2, 6,10, 14
    If d = -2 numers are 14,10,16,2

    Question 357
    CBSEENMA10009718
    Wired Faculty App

    The common difference of AP 13q. 1-6q3q, 1-12q3q, ........ is;1 1 6q 1 12q, , 3q 3q 3q
    ... is:

    • q

    • -q

    • -2

    • 2
    Solution

    C.

    -2

     Comman difference = 1 - 6q3q - 13q = 1-6q-13q = -6q3q = -2

    Question 358
    CBSEENMA10009719
    Wired Faculty App

    How many three-digit natural numbers are divisible by 7?
    Solution

    Three digit numbers divisible by 7 are 

    105, 112, 119, .......994

    This is an AP with first term (a) =105 and comman difference (d)= 7

    an be the last term.an  = a+ (n - 1 )994 = 105 + (n - 1 ) ( 7 )7 ( n - 1 ) =889n - 1 = 127n = 128Thus, there are 128 three-digit natural terms that are divisible by 7.

    Question 359
    CBSEENMA10009725
    Wired Faculty App

    The sum of first n terms of an AP is 3n2 + 4n. Find the 25th term of this AP.
    Solution

    Sn =3n2 + 4nFirst term ( a1 ) = S1 = 3 ( 1 )2 + 4 ( 1 ) = 7S2 = a1 + a2 = 3 ( 2 )2 + 4 ( 2 ) = 20a2 = 20 - a1 = 20 - 7 = 13So, comman difference ( d ) = a2 - a1 =13 - 7 =6Now,  an = a + ( n- 1 )d∴ a25 = 7 + ( 25 - 1 ) x 6            = 7 + 24 x 6            = 7 + 144            = 151

    Question 360
    CBSEENMA10009731
    Wired Faculty App

    Find the number of terms of the AP -12, -9, -6,... 12. If 1 is added toeach term of this AP, then find the sum of all terms of the AP thusobtained.
    Solution

    Given A.P., is  -12, -9, -6, ......,21

    first term, a= -12

    comman difference, d=3

    Let 21 be the nth  of the A.P.21=a+(n-1)d⇒ 21 -12 +(n-1)x3⇒ 33= (n-1)x3⇒ n=12Sum of the terms of the A.P. = S12  = n2 (2a + (n-1) d  )  = 122( -24 + 11 x 3)  = 54If 1 is added to each term of AP, the sum of all the terms of the new AP will increase by n, i.e.,12.∴ sum of all the terms of the new AP = 54+12=66

    Question 361
    CBSEENMA10009750
    Wired Faculty App

    If k, 2k- 1 and 2k + 1 are three consecutive terms of an A.P., the value of k is

    • 2

    • 3

    • -3

    • 5
    Solution

    B.

    3

    K,  2K-1,  2K+1  are in Arithmetic Progression2x(2k-1) = k+2k+14k-2=3k+1k=3

    Question 362
    CBSEENMA10009757
    Wired Faculty App

    Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
    Solution

    Number which are divisible by both 2 and 5 are the number which are divisible by 10.

    Thus we need to find the number of natural numbers between 101 and 999 which are divisible by 10.

    The first number between 101 and 999 which is divisible by 10 is 110.

    And the last number between 101 and 999 which is divisible by 10 is 990.

    Using the formula for Arithmetic progression where first term (a) = 110, and last term (Tn) = 990 and difference (d) =10.

                 Tn = a+(n-1) x d

               990 = 110+(n-1) x 10 

               880 = (n-1) x 10

                88 = n-1

                  n = 89

      Hence there are 89 natural numbers between 101 and 999 which are divisible by both 2 and 5.

    Question 363
    CBSEENMA10009764
    Wired Faculty App

    The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP.
    Solution

    The sum of the  2nd and the 7th terms of an AP is 30

    a + d + a + 6d = 302a + 7d = 3015 th term is 1 less than twice the 8th terma + 14d = 2 (a + 7d ) -1a + 14d = 2a + 14d -1a = 1Now,  2 x 1 + 7d = 30d = 4AP; 1, 5, 9,.................

    Question 364
    CBSEENMA10009772
    Wired Faculty App

    In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1to 12 classes in the school and each class has two sections, find how many trees were planted by the students. Which value is shown in this question?
    Solution

    Class 1 plant trees = 2 x class 1 x 2 section

                                = 2 x 1 x 2 = 4 x class

                                = 4 x 1 = 4 trees

    class 2 plant trees = 4 x class

                               = 4 x 2  = 8 trees

                            a =  4

                            d =  8

                            n =  12

       Sn = n22a + (n - 1) dSn = 122 2 x 4 +(12 - 1)  x 4Sn = 6 x( 8 + 11 x 4)Sn = 6 x ( 8 + 44)Sn = 6 x 52Sn = 312 trees 

    Question 365
    CBSEENMA10009782
    Wired Faculty App

    What is the common difference of an A.P. in which a21- a7= 84?
    Solution

    Let a be the first term and d be the common difference of the given A.P.

    ∴a21 - a7 = 84⇒a + 20d - a + 6d = 84⇒a + 20d -a - 6d = 84 ⇒14d = 84 ⇒d = 6Hence, the common difference is 6.

    Question 366
    CBSEENMA10009788
    Wired Faculty App

    For what value of n, are the nth terms of two A.Ps.  63, 65, 67,…. and 3, 10, 17,….. equal?
    Solution

    For A.P.  63, 65, 67,........, we have

    First term = 63  and common difference = 65-63 = 2

    Hence, nth term = an = 63 + (n-1) 2

    ⇒an = 63 + 2n - 2 = 2n + 61

     

    For A.P.  3, 10, 17,........, we have 

    First term = 3 and  common difference = 10-3 = 7

    Hence,  ntn term = an = 3 + (n-1) 7

    ⇒an = 3 + 7n - 7 = 7n - 4The two A.Ps. will have identical nth term, if an = an ⇒2n + 61 = 7n - 4⇒5n = 65⇒n = 13

    Question 367
    CBSEENMA10009796
    Wired Faculty App

    How many terms of an A.P. 9, 17, 25, …. must be taken to give a sum of 636?
    Solution

    Let there be n terms of this A.P.

    For this A.P.,  a = 9

    d = a2 - a1 = 17 - 9 = 8

       sn = n22a + (n-1) d⇒ 636 =  n22 x 9 + (n-1) x 8⇒ 636 = n9 + (n-1) x 4⇒ 636 = n ( 9 + 4n - 4 ) ⇒ 636 = n( 4n + 5 )⇒ 636 = 4n2 + 5n⇒ 4n2 + 5n - 636 = 0⇒   4n2 + 53n - 48n - 636 = 0⇒ n ( 4n + 53 ) - 12 ( 4n + 53 ) = 0⇒ ( 4n + 53 )  ( n - 12 ) = 0⇒ 4n + 53 = 0  or  n-12 = 0⇒  n = -534   or  n= 12

    Since number of terms can neither be negative nor fractional,

    we have  n = 12.

    Question 368
    CBSEENMA10009804
    Wired Faculty App

    If the ratio of the sum of the first n terms of two A.Ps is  (7n + 1) : (4n + 27), then find the ratio of their 9th terms.
    Solution

    Let a1, a2 be the first terms and d1, d2 the common difference of the two given A.P.'s 

    Then, sum of their n terms is given by

    Sn = n22a1 + (n-1) d1    and   Sn = n22a2 + (n-1) d2∴  snsn =  n22a1 + (n-1) d1 n22a2 + (n-1) d2                       =  2a1 + (n-1) d12a2 + (n-1) d2

    It is given that, 

    snsn = 7n + 14n + 27⇒ 2a1 + (n-1)d12a2 + (n-1)d2 = 7n + 14n + 27  ..........(i)

    In order to find the ratio of the mth terms of the two given A,P.'s,

    We replace  n  by [ 2m-1] in equation (i)

     

    2a1 + (17-1)d12a2 +(17-1)d2 = 7 x 17 + 14 x 17 + 27⇒2a1 + 16d12a2 +16d2 = 2419⇒a1 + 8d1a2 +8d2 = 2419

    Thus, the ratio of their 9th terms is  24:19.

     

    Question 369
    CBSEENMA10009823
    Wired Faculty App

    Find the sum of all three digit natural numbers, which are multiples of 7.
    Solution

    First three-digit number that is divisible by 7 = 105

    Next number = 105 + 7 = 112

    Therefore the series is  105, 112, 119,........

    The maximun possible three digit number is 999.

    When we divide by 7, the remainder will be 5.

    Clearly, 999 - 5 = 994 is the maximum possible three-digit number divisible by 7.

    The series is as follows:

    105, 112, 119,.........994.

     Here  a = 105,  and  d = 7

    Let 994 be the nth term of this A.P.

          an = a + ( n-1 ) d

    ⇒ 994 = 105 + ( n-1 ) x 7

    ⇒  ( n-1 ) x 7 = 889

    ⇒  ( n-1 ) = 127

    ⇒   n = 128

    So, there are 128 terms in the A.P.

    ∴ Sn = n2First term + last term                     = 1282 a1 + a128           = 64 ( 105 + 994 )           = 64 x 1099           = 70336

    Question 370
    CBSEENMA10009833
    Wired Faculty App

    The 16th term of an AP is 1 more than twice its 8th term. If the 12th term of the AP is 47, then find its nth term.
    Solution

    Let a and d respectively be the first term and the common difference of the A.P.

    we know that the nth term of an AP is given by  an = a + (n-1) d

    According to the given information,

    a16 = 1 + 2a8

    ⇒  a + (1 6- 1) d = 1 + 2[a + (8 - 1) d]

    ⇒   a + 15d = 1 + 2a + 14d

    ⇒  -a + d = 1                     ........(1) 

    Also, it is given that, a12 = 47

    ⇒  a + ( 12 - 1 ) d = 47

    ⇒  a + 11d = 47                 .......(2)

    Adding  (1) and (2), we have:

    12d = 48

    ⇒  d = 4

    From (1),

    -a + 4 = 1  

    ⇒  a = 3

    Hence,  an = a + (n - 1) d

                    = 3 + (n - 1) (4)

                    = 3 + 4n - 4

                    = 4n - 1  

    Hence, the nth term of the AP is  4n - 1.

    Question 371
    CBSEENMA10009839
    Wired Faculty App

    Sum of the first 20 terms of an AP is -240, and its first term is 7. Find its 24th term.
    Solution

    Given: S20 = -240  and a = 7

    Consider,  S20 = -240

       Sn = n22a + (n-1) d⇒  -240 =  2022 x 7 + (20-1) d⇒ -240 = 10 (14 + 19d)⇒ -24  = 14 + 19d ⇒  -24 - 14 = 19d⇒  19d = -38⇒ d= -2

    Now, a24 = a + 23d

                  = 7 + 23 x (-2)

                  = -39

    Hence, a24 = -39

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