Arithmetic Progressions

  • Question 1
    CBSEENMA10007384

    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 = aTaxi fare for 2 kms
    = Rs. 15 + Rs. 8 = Rs. 23 = aTaxi fare for 3 kms
    = 23 + Rs. 8 = Rs. 31 = aTaxi fare for 4 kms
    = Rs. 31 + Rs. 8 = Rs. 39 = a4
    and so on.
    a– 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

    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

    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 = aTaxi fare for 2 kms
    = Rs. 15 + Rs. 8 = Rs. 23 = aTaxi fare for 3 kms
    = 23 + Rs. 8 = Rs. 31 = aTaxi fare for 4 kms
    = Rs. 31 + Rs. 8 = Rs. 39 = a4
    and so on.
    a– 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

    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
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    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.

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