Sponsor Area
Write the first five terms of each of the sequences whose nth terms are :
Here
Putting n = 1, 2, 3, 4, 5, we get
∴ First five terms are
Write the first five terms of the sequence whose nth terms are:
Here,
Putting n = 1, 2, 3, 4, 5, we get
∴ First five terms are: 25, -125, 625, -3125, 15625
Find the indicated terms in the following sequence whose nth terms are :
Here,
Putting n = 19, we get
Find the indicated terms in the following sequence whose nth terms are:
Here, ...(i)
Replacing n by n-1 in (1), we get
Putting n = 16 in (I), we get
The Fibonacci sequence is defined by
Find , for n = 1, 2, 3, 4, 5.
The Fibonacci sequence is given by
and
Putting n =3 in (i), we get
Putting n = 4 in (i), we get
Putting n = 5 in (i), we get
Putting n = 6 in (i), we get
Now, let
Find the first six terms of the sequence whose first term is 1 and whose (n+l)th term is obtained by adding n to the nth term.
Let denote the nth term
According to given equation ...(i)
Putting n = 1 in (i), we get
Putting n = 2 in (i), we get
Putting n = 3 in (i), we get
Putting n = 4 in (i), we get
Putting n = 5 in (i), we get
Consider the sequence defined by tn = an2 + bn + c. If t2 = 3, t4 = 13 and t7 = 113, show that 3tn = 17n2 – 87n + 115
Here, ...(i)
Now,
4a + 2b + c = 3 ...(ii)
16a + 4b + c = 13 ...(iii)
49a + 7b + c = 113 ...(iv)
Subtracting (iii) from (iv), we get 33a + 3b = 100 ...(v)
Subtracting (ii) from (iii), we get 12a + 2b = 10 6a + b = 5 ...(vi)
Multiplying both sides of (vi) by 3, we get
18a + 3b = 15 ...(vii)
Subtracting (vii) from (v), we get
Using this value of a in (vi), we get
Using values of a and b in (ii), we get
Using values of a, b, c in (i), we get
Write first five terms of the following sequence and obtain the corresponding series. for all n>1
Here,
Now, ...(i) (for n>1)
Putting n = 2 in (i), we get
Putting n = 3 in (i), we get
Putting n =4 in (i), we get
Putting n = 5 in (i), we get
The corresponding series is : 3 + 11 + 35 + 107 + 323 + ..........
Write first five terms of the following sequence and obtain the corresponding series., n>2
Here,
...(i) (for n>2)
Putting n = 3 in (i), we get
Putting n = 4 in (i), we get
Putting n = 5 in (i), we get
The corresponding series is: 2 + 2 + 1 + 0 - 1 ...............
Show that the sequence defined by
(where A and B are constant) is an A.P. with common difference A.
Here,
Replacing n by n - 1, we get
which is constant and independent of n.
Hence, the sequence is an A.P.
Show that the sequence defined by
is not an A.P.
Here,
Replacing by n - 1, we get
(which depends upon n and is therefore not a constant)
Hence, the given sequence is not an A.P.
Which term of an A.P. 5, 2, -1, ......... is -22?
The given A.P. is: 5, 2, - 1,...............
Here a = 5, d = 2 - 5 = -3
Let -22 be the nth term
∴ or a + (n - 1)d = -22
or 5 + (n - 1) (-3) = -22 or 5 - 3n + 3 = -22
or 8 - 3n = -22 or 3n = 8 + 22
or 3n = 30 or n = 10
∴ -22 is the 10th term.
Which term of the sequence 12 + 8i, 11 + 6i, 10+41i, ....... is
(i) purely real (ii) purely imaginary?
The given sequence is :
12 + 8i, 11 + 6i, 10+4i, .......
Here, a = 12 + 8i
d = (11 + 6i) - (12 + 8i) = -1 - 2i
(i) Let nth term be purely real
∴ is purely real
or 12 + 8i + (n - 1) (-1 - 2i) is purely real
or 12 + 8i - (n - 1) - 2 (n - 1) is purely real
or (13 - n) + (10 - 2n) i is purely real
∴ 10 - 2n = 0 or n = 5
∴ 5th term is purely real.
(ii) Let nth term be purely imaginary
∴ is purely imaginary
or (13 - n) + (10 - 2n) is purely imaginary
or 13 - n = 0
or n = 13
∴ 13th term is purely imaginary.
Which term of the sequence is the first negative term?
The given sequence is an A.P. with a = 24,
Let the nth term of the given A.P. be the first negative term.
Then,
or 24 + (n - 1) or 96 - 3n + 3 < 0
or 99 - 3n < 0 or 3n > 99
or n > 33
Since 34 is the natural number just greater than 33.
∴ n = 34, Hence, 34th term of the given A.P. is the first negative term.
Determine the number of terms in an A.P. 3, 8, 13, .......253. Also, find 12th term from the end.
The given A.P. is : 3, 8, 13, ....., 253
Here, a = 3, d = 8 - 3 = 5
Let there be n terms of an A.P.
∴ or
or a + (n - 1)d = 253 or 3 + (n - 1) 5 = 253
or 3 + 5n - 5 = 253 or 5n - 2 = 253
or 5n = 255 or n = 51
Since nth term from end =
∴ 12th term from end = 253 - (12 - 1)5 = 253 - (11) (5) = 253 - 55 = 198
Find the 2nd term and the rth term of an A.P. whose 6th term is 12 and 8th term is 22.
Let a be the first and d be the common difference
...(i)
... (ii)
Subtracting (ii) from (i), we get
-2d = -10 or d = 5
using value of d in (i), we get
a + 5(5) = 12 or a = 12 - 25 = -13
Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.
Since K + 2, 4K - 6 and 3K - 2 are three consecutive terms of an A.P.
∴ (4K - 6) - (K + 2) = (3K - 2) - (4K - 6)
or 4K - 6 - K - 2 = 3K - 2 - 4K + 6 or 3K - 8 = -K + 4
or 4K = 12 or K = 3
Sponsor Area
In an A.P., the mth term is and the nth term is
find the (mn)th term.
Let a be the first term and d be the c.d.
...(i)
... (ii)
Subtracting (ii) from (i), we get
(m - 1)d - (n - 1)d = or (m - 1 - n + 1)d =
or or
Using this value of d in (i), we get
or
or or
If m times the mth term is equal to n times the nth term of an A.P. Prove that (m + w)th term of an A.P. is zero.
Let a be the first term and d be the common difference
Since m times the mth term is equal to n times the nth term.
∴
or m[a + (m - 1)d] = n[a + (n - 1)d]
or ma + m(m - 1)d = na + n (n-1)d
or ma - na +
or (m - n) a + (
or (m - n) a + [(m2 - n2) - (m - n)] d = 0
or (m - n) a + [(m - n) (m + n) - (m - n)]d = 0
or (m - n) a + (m - n) (m + n - 1)d = 0
or (m - n) [a + (m + n - 1)d] = 0
or a + (m + n - 1) d =0 (∵ )
Hence,
If (m + l)th term of an A.P. is twice the (n + l)th term. Prove that (3m + l)th term is twice the (m + n + l)th term.
Let a be the first and d be the common difference of an A.P. Since (m + 1)th term of an A.P. is twice the (n + 1)th term.
∴
or a + (m + 1 - 1)d = 2[a + (n + 1 - 1)d]
or a + md = 2a + 2nd
or a = (m - 2n)d
Now,
...(i)
Now,
or ...(ii)
∴ (3m + 1)th term is twice the (m + n + 1)th term of an A.P.
The pth, qth and rth terms of an A.P. are a, b, c respectively. Show that (q – r) a + (r – p) b + (p – q)c = 0
Let A be the first term and D be the c.d. of an A.P.
...(i)
...(ii)
...(iii)
L.H.S. = (q - r) a + (r - p)b + (p - q)c = (q - r) [A + (p - 1)D]
+ (r - p) [A + (q - 1)D + (p - q) [A + (r - 1)D]
(By using values of a, b, c from (i), (ii), (iii) respectively]
= A(q - r) + (q - r) (p - 1)D + A (r - p) + (r - p) (q - 1)D + A(p - q) + (p - q) (r - 1)D
= A(q - r + r - p + p - q) + D[Pq - q - rp + r + rq - r - pq + p + pr - p - qr + q]
= A(0) + D(0) = 0 = R.H.S.
Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Let a be the first term and d be the common difference
∴ ...(i)
and ...(ii)
Adding (i) and (ii), we get
=
or
Hence, the sum of and
terms is twice the mth term.
Find the sum to n terms of an A.P., whose nth term is
Here, ...(i)
Replacing n by (n - 1), we get
...(ii)
∴ d =
∴ The sequence is an A.P.
a = -1, d = -6
Solve the equation 1 + 6 + 11 + 16 + ........ + x = 148
The given equation is : 1 + 6 + 11 + 16 +...........+x = 148
Here, a = 1, d = 6 -1 = 5, l = x, Let x be the nth term
∴
or
or
or n[2 + 5n - 5] = 296
or n (5n - 3) = 296
or
or n = 8 (∵ n is a natural number)
∴ or a + (8 -1)d = x or 1+7 (5) = x or x = 36
How many terms of the A.P. -6, -5, .......... are needed to give the sum -25? Explain the double answer.
The given A.P. is: -6, -5, .............
Here, a = -6,
Let -25 be the sum of n terms
∴ or
or or
or or n(n - 25) = -100
or
or
Here, the common difference is positive.
∴ The A.P. starts from negative terms and its terms are increasing.
∴ All the terms after 13th term are positive.
These positive terms from 14th term to 20th term when added to negative terms from 6th term to 12th term, they cancel out each other and the sum remains same.
Hence, the sum of first five terms is same as the sum of first 20 terms.
In a A.P., the first term is 2 and the sum of first five terms is one-fourth of the sum of the next five terms. Show that the 20th term is – 112.
Here, a = 2, Let d be the common difference. Since the sum of first five terms is one-fourth of the sum of the next five terms.
or or
or
or
or or 50 + 50d = 20 + 45d
or 5d = -30 or d = -6
Now,
= 2 + 19 (-6) = 2 - 114 = -112
If the sum of first n terms of a sequence is of the form An2 + Bw, where A and B arc constants (independent of n). Show that the sequence is an A.P. Is the converse true? Justify your answer.
Here, ...(i)
Replacing n by (n - 1) in (i), we get
= ...(ii)
=
Replacing n by (n - 1) in (iii), we get
Now,
which is independent of n
∴ The sequence is an A.P.
Yes, the converse is true.
Let a be the first term and d be the common difference.
∴
where and
are fixed numbers.
Hence, if the sequence is an A.P. then the sum of first n terms must be of the form
In an A.P., if the pth term is and the qth term is
prove that the sum of the first pq terms must be
where
Let a be the first term and d be the common difference.
...(i)
...(ii)
Subtracting (ii) from (i), we get or
or or
Using this value in (i), we get
or
or or
If the pth term of an A.P. is a and qth term is b, show that the sum of (p + q) terms
is
Let A be the first term and D be the common difference.
...(i)
...(ii)
Subtracting (ii) from (i), we get
(p - 1)D - (q - 1) D = a - b or (p - q)D = a - b
or ...(iii)
Adding (i) and (ii), we get
2A + (p - 1)D + (q - 1) D = a + b or 2A + (p - 1 + q - 1)D = a + b
or 2A + (p + q - 1)D - D = a + b (Note this step carefully)
or 2A + (p + q - 1)D = a + b + D
or 2A + (p + q - 1)D = a + b + { By using (iii)} ...(iv)
[By using (iv)]
Find the sum of odd integers from 1 to 2001.
All the odd integers from 1 to 2001 are 1, 3, 5 , 7 , 9,.............. 2001.
Let denote their sum, i.e.
Here, a = 1, d = 2,
i.e., a + (n - 1) d = 2001 or 1 + (n - 1)2 = 2001
or 1 + 2n - 2 = 2001 or 2n = 2001 + 1 = 2002
or
∴ [∵
]
Find the sum of all natural numbers lying between 100 and 1000, which are mutliples of 5.
All the natural numbers between 100 and 1000 which are multiples of 5 are:
105, 110, 115, ......, 995.
Let denote their sum, i.e.,
Here, a = 105, d = 110 - 105 = 5,
i.e., a + (n - 1)d = 995 or 105 + (n - 1)5 = 995
or 5n + 100 = 995 or 5n = 995 - 100 = 895
or
∴
Let the sum of n, 2n, 3n terms of an A.P., be S1, S2, S3, respectively, show that
Let a be the first term and d be the common difference.
∴ ...(i)
...(ii)
... (iii)
Subtracting (i) from (ii), we get
[B using (3)]
Hence,
Find the sum of all the numbers between 200 and 400 which are divisible by 7.
All the numbers between 200 and 400, which are divisible by 7 are:
203, 210, 217 ......, 399.
Here, a = 203, d = 210 - 203 = 7,
or a + (n - 1)d = 399
203 + (n - 1) 7 = 399 (n - 1) 7 = 399 - 203 = 196
∴ [Use
]
=
Find the sum of all the two digit numbers, which when divided by 4, yields 1 as remainder.
All the numbers, which consist of two digits and leave 1 as remainder when divided by 4 are: 13, 17, 21, ........., 97.
Here, a = 13, d = 4, a + (n - 1)d = 97
13 + (n - 1) 4 = 97
(n - 1) 4 = 97 - 13 = 84
n - 1 =
n = 22
∴
Find sum of integers from 1 to 100 that are divisible by 2 or 5.
Let = Sum of all the numbers from 1 to 100 which are divisible by 2.
= Sum of all the numbers from 1 to 100 which are divisible by 5.
= Sum of all numbers from 1 to 100 which are divisible by both 2 and 5.
∴ Required Sum = ...(i)
Now,
Here,
a + (n - 1)d = 1800
2 + (n - 1)2 = 100
or (n - 1)2 = 98 or n - 1 = or n = 50
∴
Here, a = 5, d = 5, a + (n - 1)d = 100 or 5 + (n - 1)5 = 100
n - 1 = 19
n = 20
∴
∴ from (i), we have
Required sum = 2550 + 1050 - 550 = 3050.
The sum of first 4 terms of an A.P. is 56. The sum of last 4 terms is 112. If its first term is 11, then find the number of terms.
Let a be first term, d be common difference, l be the last term of A.P.
Given a = 11,
22 + 3d = 28
3d = 6
d = 2.
Also, Sum of last 4 terms = 112l + (l -d) + (l - 2d) + (l - 3d) = 112
41 - 6d = 112
Let 31 be the nth term.
∴
or 11 + (n - 1) 2 = 31 or (n - 1)2 = 20 or n - 1 = 10 or n = 11.
The sums of first n terms of two A.P.’s are in the ratio of 14 – 4n : 3n + 5. Find the ratio of their 8th terms.
Let and
respectively be the first term and common difference of two A.P.'s
or
or or
Taking or n - 1 = 14 or n = 15
∴
The ratio of sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of with and nth term is 2m – 1 :2n – 1.
Let a be the first term and d be the common ratio. or
or or 2an + n (m - 1)d = 2 am + m(n - 1)d
or 2am - 2an = (mn - n) d - (mn - m)d
or 2a (m - n) = d [mn -n -mn + m]
or 2a(m - n) = d (m -n) or d = 2a
∴ The ratio of mth term and nth term is:
2m - 1 : 2n - 1
If S1, S2, S3 ........Sp are the sums of n terms of p A.P.’s whose first terms are 1, 2,3, .....p and common differences are 1, 3, 5, .....(2p – 1) respectively. Show that
Here, = Sum of first n terms of an A.P. with first term 1 and c.d. 1.
=
Sum of first n terms of an A.P. with first term 2 and c.d. 3.
=
= Sum of first n terms of an A.P. with first term 3 and c.d. 5.
=
.............................................
..............................................
Sum of first n terms of an A.P. with first term p and c.d. (2p - 1)
∴ =
=
=
=
Sponsor Area
Insert 6 arithmetic means between 3 and 24.
Let be six A.M.'s between 3 and 24.
∴ 3, are in A.P.
Here, a =3, Let d be the common difference
∴ Six A.M.'s between 3 and 24 are 6, 9, 12, 15, 18, 21.
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of the 7th and (m – l)th numbers is 5 : 9. Find the value of m.
Let be m A.M.'s between 1 and 31.
∴ 1, are in A.P.
Here, a = 1,
Let d be the common difference or a + (m+ 2 - 1)d = 31 or 1 + (m + 1)d = 31 or
Now, or
or or 9m + 1899 = 155m - 145
or 146m = 2044 or m = 14
Hence, m = 14
For what value of n, is the A.M. between a and b?
Since A.M. between a and b is
∴ or
or or
or or
or or
or or
or n = 0
Prove that the sum of n arithmetic means between two numbers is n times the single A.M. between them.
Let a and b be two given numbers.
Let be A.M.'s between a and b.
∴ a, b are in A.P.
{∵ First term = a, last term = b and number of terms = n + 2}
= n times the single A.M. between a and b.
The sum of two numbers is . An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Find the number of means inserted.
Let a and b be two numbers such that
...(i)
Let be 2n (even) A.M.'s between a and b.
∴ (By using example 4)
=
Also,
∴ or 12n + 6 = 13n or n = 6
Hence, 12 A.M.'s are inserted.
Insert A.M.’s between 7 and 71 in such a way that the 5th A.M. is 27. Find the number of A.M.’s.
Let be n A.M.'s inserted between 7 and 71.
∴ 7, are in A.P.
a = 7,
Let d be the common difference
a + (n + 2 - 1)d = 71
7 + (n + 1)d = 71
...(i)
Now, or
or a + 5d = 27 or 7 + 5d = 27
or 5d = 20 or d = 4 ...(ii)
From (i) and (ii), we have
or n + 1 = 16
or n = 15
∴ 15 A.M.'s are inserted between 7 and 11.
The sum of three numbers in A.P. is 21 and the product of first and the third numbers exceeds the second number by 6, find the three numbers.
Let the three numbers in A.P. be a - d, a, a + d.
Since the sum of the number = 21
∴ a - d + a + a + d = 21 or 3a = 21
or a = 7
Since the product of first and the third numbers exceeds the second number by 6.
∴ (a - d) (a + d) = a + 6 or
or or 49 - 13 =
or or
When d = 6, numbers are 7 - 6, 7, 7 + 6
or 1, 7 , 13
When d = -6, numbers are 7 + 6, 7, 7 - 6
or 13, 7, 1
Find four numbers in A.P. whose sum is 32 and the product of extremes is to the product of means is 7 : 15.
Let the four number in A.P. be a - 3d, a - d, a + d, a + 3d.
Since the sum of four numbers = 32
∴ a - 3d + a - d + a + d + a + 3d = 32
or 4a = 32 or a = 8
Since the ratio of the product of extremes to the product of means is 7:15.
∴ or
or or
or or
or or
When d = 2 and a = 8 numbers are 8 - 3(2), 8 - 2, 8 + 2, 8 + 3(2)
2, 6, 10, 14
When d = -2 and a = 8 numbers are 8 - 3(-2), 8 - (-2), 8 + (-2), 8 + 3(-2)
14, 10, 6, 2
If the lengths of the sides of a right angled triangle are in A.P., then show that their ratio is 3 : 4 : 5.
Let a - d, a, a + d, with d>0 be the lengths of sides of a right angled triangle.
∴ a + d, the largest side must be hypotenuse.
So, by pythagorus theorem
or
or or a = 4d
∴ Sides are 4d - d, 4d, 4d + d or 3d, 4d, 5d
Hence, the sides are in the ratio 3:4:5.
The digits of a positive integer having three digits are in A.P. and their sum is 15.
The number obtained by reversing the digits is 594 less than the original number.
Find the number.
Let a - d, a, a + d be the three digits of a three digit number respectively at unit place, ten's place, hundreth place.
∴ The number = 100 (a + d) + 10 + a + a - d
Since the sum of digits is 15
∴ a -d + a + a + d = 15 or 3a = 15 or a = 5
The number obtained by reversing the digits = 100 (a - d) + 10a + (a + d) ...(ii)
According to the given condition
100 (a - d) + 10 a + a + d = 100 (a + d) + 10a + a - d - 594
or 100a - 100d + 10a + a + d = 100a + 100d + 10a + a - d - 594
or 198d = 594 or d = 3
∴ Original number = 100 (5 + 3) + 10 (5) + 5 - 3 = 800 + 50 + 2 = 852
Verification: On reversing the digits, the number is 258
852 - 258 = 594
If and
are in A.P., Prove that
are also in A.P.
Here, are in A.P.
or are in A.P. (By adding 1 to each other)
or are in A.P.
or are in A.P. {By dividing each number by (a + b + c) (
)}
If are in A.P. prove that a, b, c are in A.P.
If a, b, c are in A.P., prove that are also in A.P.
If are in A.P., then prove that
are in A.P.
If are in A.P., show that
are in A.P.
If are in A.P., prove that
are in A.P.
Since are in A.P.
∴
or
or (b - a) (2c - a - b) = (c - b) (2a - b - c) ...(i)
Now, will be in A.P.
if or if
=
or if (Note this step carefully)
or if (b - a) (2c - a - b) = (c - b) (2a - b - c) which is true. [By using (i)]
Hence, are in A.P.
If a, b, c are in A.P. then prove that
Since a, b, c are in A.P.
∴ b - a = c - b or 2b = a + c or ...(i)
if
[ By using (i)]
if or if
if - 2ac = -2ac which is true
∴
If a, b, c are in A.P. then prove that
Since a, b, c are in A.P.
∴ b - a = c - b or 2b = a + c or b = ...(i)
if [BY using (i)]
if
if
if which is true
∴
Find the 10th term of the series 5 + 25 + 125 + ....... . Also, find its wth term.
The given series is:
5 + 25 + 125 + .........., It is a G.P.
Here,
10th term =
nth term =
which term of the series:
The given series is:
It is a G.P. Here,
Let be the nth term
∴
or
or
or n - 1 = 7 or n= 8
∴ is the 8th term.
The fourth term of a GP. is 27 and the 7th term is 729, find the GP.
Let a be the first term and r be the common ratio
...(i)
...(ii)
Dividing (ii) by (i), we get
Using r = 3 in (i), we get
∴ G.P. is 1, 1(3), 1(3)2, 1(3)3, ........... or 1, 3, 9, 27
The seventh term of a GP. is 8 times the fourth term and 5th term is 48. Find the GP.
Let a be the first term and r be the common ratio.
According to the given condition,
Also,
∴ G.P is 3, 3(2), 3(2)2, ....... or 3, 6, 12,.......
If the GP.’s 5,10, 20, .......and 1280, 640, 320, ......have their nth terms equal, find the value of n.
The first G.P. is : 5, 10, 20,...........
Here, a = 5,
nth term =
The second G.P. is: 1280, 640, 320, ...........
Here, a = 1280,
nth term =
Since the nth terms of both the G.P.'s are equal
∴
The third term of a GP. is 4. Find the product of its first five terms.
Let a be the first term and r be the common ratio
...(i)
Product of first five terms =
= {By using (i)}
The 5th, 8th and 11th terms of a GP. are p, q and s respectively. Show that q2 = ps.
Let a be the first term and r be the common ratio
Now, ...(i)
...(ii)
From (i) and (ii), we have
Find all the sequences which are simultaneously in A.P. and GP.
Let the sequence be in A.P. as well as in G.P.
Let be three consecutive terms of an A.P.
∴
or ...(i)
Let r be the common ratio of G.P.
∴
and
∴
Hence, all the constant sequences are the only sequences which are in A.P. as well as in G.P.
In a finite GP. the product of the terms equidistant from the beginning and the end is always same and equal to the product of first and last term.
Let be a finite G.P. with common ratio r. Then,
= mth term from the beginning =
Now,
If the first and the nth terms of a GP. are a and b respectively and if P is the product of first n terms, prove that P2 = (ab)n.
Here, a is the first term, let r be the common ratio.
Also,
Since P is the product of first n terms
∴
Squaring both sides, we get
If show that a, b, c, d are in G.P.
Given (a + bx) (b - cx) = (a - bx) (b + cx)
or ab - acx + b2x - bcx2 = ab + cax - b2x - bcx2
or b2x + b2x = cax + acx 2b2x = 2acx or b2 = ac or
...(i)
Also,
or or
...(ii)
From (i) and (ii), we have
Hence, a, b, c are in G.P.
If 4th, 10th and 16th terms of a GP. are x, y and z respectively, Prove that y, z are in GP.
Let a be the first term and r be the common ratio of G.P.
Given
...(i)
...(ii)
...(iii)
Multiply (i) and (iii), we get
or or
[By using (ii)]
∴ or
or x, y, z are in G.P.
Show that products of the corresponding terms of sequences a, ar, ar2, .........ar n – 1 and A, AR, AR2, ........ An – 1 form a GP. and find common ratio.
If we multiply corresponding terms of given two sequences.
we get
Here,
∴ Above sequence is a G.P. with common ratio r . R.
The first term of GP. is 1. The sum of 3rd and 5th term is 90. Find the common ratio of G.P.
Let a be first term and r be the common ratio of G.P.
Given a = 1, also,
(∵ a = 1)
or
or
Find the sum of the indicated terms of each of the following CP’s:
3, 6, 12, ........; 7 terms
The given sequence is: 3, 6, 12,........
Here, a = 3,
Find the sum of the indicated terms of each of the following CP’s:
The given G.P. is:
Here,
Find the sum to indicated number of terms in each of the geometric progressions:
1, - a, n terms
The given G.P. is 1, -a, a2, -a3, ...........
Here, first term A = 1,
Find the sum to indicated number of terms in each of the geometric progressions:
The given G.P. is:
or
Here, and
(∵ r < 1)
Find the sum of n terms of sequence
Let denote the sum of first n terms of the given sequence.
∴
=
=
Prove that the sum to n terms of the series
11 + 103 + 1005 +........ is
Let Sn = 11 + 103 + 1005 + .......... up to n terms
= (10 + 1) + (100 + 3) + (1000 + 5) + ............... upto n terms
= [10 + 100 + 1000 + ..... upto n terms] + [1 + 3 + 5 + .......upto n terms]
= [10 + 102 + 103 + ....... upto n terms] + [ 1 + 3 + 5 +........... upto n terms]
Since the series in the first bracket is a G.P. series with first term 10, common ratio 10 and the
series in the second bracket is an A.P. series with first term 1 and common difference 2.
∴
Find the sum to n terms of the sequence given by
The given sequence is defined by
Let denote the sum of n terms
∴
=
Since the series in the first bracket is a G.P. series with 1st term 2, common ratio 2 and the series in the second bracket is an A.P. with 1st term 3, common difference 3.
∴
Sponsor Area
How many terms of the sequence must be taken to make the sum
?
The given sequence is:
Here,
Let be the sum of n terms
∴ or
or or
or
or or
or or
or n/2 = 3 or n = 6
Sum the following series:
0.5 + 0.55 + 0.555.......... to n terms
Let = 0.5 + 0.55 + 0.555 +............ to n terms
= 5[0.1 + 0.11 + 0.111 + ........... to n terms]
=
=
=
=
Given a G.P. with a = 729 and 7th term 64, determine S7.
Here a = 729, Let r be the common ratio
When then
When then
Determine the number of terms in a GP., if t1 = 3, tn = 96 and Sn = 189.
t1 = 3 first term a =3
Let r be common ratio
3
32r - 1 = 63r - 63
31r = 62
r = 2
Now,
n - 1 = 5
n = 6
Hence, the number of terms is 6.
Find the least value of n for which the sum 1 + 3 + 32+ ...... to n terms is greater than 7000.
Let
Here, a = 1, r = 3
Since
∴
Hence, the least value of n is 9.
The ratio of the sum of first three terms is to that of first 6 terms of a G.P. is 125 : 152. Find the common ratio.
Let a be the first term and r be the common ratio
According to the given equation,
If S1, S2 and S3 be respectively the sums of n, 2n and 3n terms of a GP., prove that S1 (S3 – S2) = (S2 – S1)2
Let a be the first term and r be the common ratio of a G.P.
...(i)
...(ii)
...(iii)
L.H.S. =
=
R.H.S. =
=
∴ L.H.S. = R.H.S.
Hence,
Let S be the sum, P the product and R the sum of the reciprocals of n terms in a G.P. Prove that
Let a be the first term and r be the common ratio of a G.P.
S = the sum of n terms of G.P. = a + ar + .........arn-1
∴ ...(i)
P = the product of n terms of G.P. = a. ar. ar2 ....... arn-1
∴ P =
...(ii)
R = the sum of the reciprocals of n terms of G.P. =
=
Now,
L.H.S. =
R.H.S. =
∴ L.H.S. = R.H.S.
Hence,
Find the sum of the products of the corresponding terms of sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2,
If we multiply the corresponding terms of given two sequences, we get 2 (128), 4 (32), 8(8), 16(2), 32 or 256, 128, 64, 32, 16 which is a G.P. with a = 256,
, n =5.
∴
If f is a function f(x + y) = f(x) f(y) for all x, y N such that f(1) = 3 and
, find the value of n.
Given f(1) = 3
and f(x + y) = f(x). f(y) ...(ii) x,
...(i)
Putting x = 1, y = 1 in (ii), we get
f(1 + 1) = f(1). f(1) f(2) = 3. 3 = 9.
Putting x = 1, y = 2 in (ii), we get
f(1 + 2) = f(1). f(2) = 3 . 9 = 27 or f(3) = 27
Putting x = 1, y = 3 in (ii), we get
f(1 + 3) = f(1). f(3) = 3. 27 = 81 or f(4) = 81
∴ f(1) = 3, f(2) = 9, f(3) = 27, f(4) = 81, ...............
Here, the sequence 3, 9, 27, 81, .................. is a G.P. with a = 3, r = 3.
Now,
or 3 + 9 + 27 +............ to n terms = 120
∴ n = 4
If A and G are respectively the A.M. and G.M. between the two distinct positive numbers are a and b, then Prove A>G.
Since A is the A.M. between a and b
∴ A = ...(i)
Since G is the G.M. between a and b
∴ G = ...(ii)
A - G =
∴ A - G > 0 A > G
If A and G are respectively the A.M. and G.M. between two positive numbers a and b, then proof the quardratic equation having a, b as it roots is
Since A is the A.M. between a and b
∴ A = ...(i)
Since G is the G.M. between a and b
∴ G = ...(ii)
From (i), a + b = 2A
∴ Sum of the roots = 2A = S (Say)
From (ii), ab = G2
∴ Product of the roots = G2 = P (say)
Required quadratic equation having a and b as its roots is
or
If A and G be A.M. and G.M. between two positive numbers, then the numbers are
Since the given numbers are the roots of the quadratic equation
∴
∴ Numbers are
The A.M. between two positive numbers is 34 and their GM. is 16. Find the numbers.
Let a and b be two numbers
A.M. = 34 ...(i)
G.M. = 16 ...(ii)
From (i), b = 68 - a
Using in (ii), we get
a(68 - a) = 256
a(a - 64) - 4 (a - 64) = 0
(a - 64) (a - 4) = 0
a = 4, 64
When a = 4, b = 68 - 4 = 64
and a = 64, b = 68 - 64 = 4
Find two positive numbers whose difference is 12 and whose A.M. exceeds the GM. by 2.
Let a and b be two positive numbers such that
a - b = 12 ...(i)
Since A.M. exceeds G.M. by 2
∴
a + b =
...(ii)
From (i), ...(iii)
Adding (ii) and (iii), we get
Subtracting (ii) from (iii), we get
Find the value of n so that may be the G.M. between a and b.
Since G.M. between a and b is
∴
The A.M. of two numbers is 3 times their G.M. Show that the numbers are in the ratio
Let a and b be two numbers.
A.M = G.M. =
According to the given condition
Applying componendo and dividendo, we get
Again applying componendo and dividendo, we get
Squaring both sides, we get
∴ Numbers are in ratio
If two G.M.'s and one A.M. 'A' be inserted between two numbers, show that
Let a and b be two numbes, since g1 and g2 are two G.M.'s between a and b.
∴ a, , b are in G.P.
Let r be common ratio
∴
Since A is the A.M. between a and b
∴ 2A = a + b ...(i)
R.H.S. = {By using (i)}
= L.H.S.
If a be the A.M. and x, y be the two G.M.'s between b and c, show that
Since a is the A.M. between b and c
∴ ...(i)
Since x, y are two G.M.'s between b and c
∴ b, x, y, c are in G.P.
Let r be the common ratio
L.H.S. =
= bc (b + c) = bc(2a) {By using (i)}
= 2abc = R.H.S.
The sum of first three terms of a G.P. is and their product is -1. Find the terms.
Let a, ar be the three numbers in G.P.
Since their products is -1
∴
or a = - 1
Since the sum of the numbers
∴
-12 - 12r - 12r2 = 13r
4r(3r + 4) + 3(3r + 4) = 0
or
When a = -1 and then numbers are
or
When a = -1 and then numbers are
or
The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is Find them.
Let the three numbers be
Since their product is 125
∴
a = 5
Since the sum of their products taken in pairs is
∴
When a = 5 and r = 2, numbers are or
When a = 5 and r = numbers are
or 10, 5,
The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.
Let three numbers in G.P. be a, ar, ar2
Since the sum of the numbers is 21
Let a, ar, ar2, ar3 be the four numbers in G.P.
Since the third number is greater than the first by 9
∴ ar2 = a + 9 a(r2 - 1) = 9 ...(i)
Since the second number is greater than the fourth by 18
∴ ar = ar3 + 18 ar(r2 - 1) = -18 ...(ii)
Dividing (ii) by (i), we get
Using r = -2 in (i), we get
a(4 - 1) = 9
a = 3
∴ Numbers are 3, 3(-2), 3(-2)2, 3(-2)3 or 3, -6, 12, -24
Let a, ar, ar2 be three numbers in G.P.
Since their sum is 56
∴ a + ar + ar2 = 56 a(1 + r + r2) = 56 ...(i)
It is given that if we subtract 1, 7, 21 from the numbers respectively then the resulting numbers form an A.P.
∴ a - 1, ar - 7, ar2 - 21 are in A.P.
...(ii)
Dividing (i) by (ii), we get
when r = 2, from (i),
∴ Numbers are 8, 8(2), 8(2)2 or 8, 16, 32
When from (i),
∴ Numbers are or 32, 16, 8
Let a - d, a, a + d be three numbers in A.P. Since their sum is 21
∴ a - d + a + a + d = 21 3a = 21
a = 7
∴ Numbers are 7 - d, 7, 7 + d
It is given that if the second number is reduced by 1 and third is increased by 1 then resulting three numbers form G.P.
∴ 7 - d, 7 - 1, 7 + d + 1 are in G.P. or 7 - d, 6, 8 + d are in G.P.
(8 + d) (7 - d) = 36
56 - 8d + 7d - d2 = 36
d =
When a = 7 and d = 4, numbers are 7 - 4, 7 + 4 or 3, 7, 11
when a = 7 and d = -5, numbers are 7 - (-5), 7, 7 - 5 or 12, 7, 2
Let A be the first term and D be the common difference.
∴
...(i)
...(ii)
...(iii)
Since are in G.P.
∴
Now, and
and
and
and
and
p - q, q - r, r - s are in G.P.
Here, a, b, c are in G.P. Let r be common ratio of G.P.
∴ b = ar, c = ar2,
Now, will be in G.P.
if or if
or if or if r = r which is true.
Hence, if a, b, c are in G.P. then are also in G.P.
Here, a, b, c, d are in G.P. Let r be the common ratio of the G.P.
∴ b = ar, c = ar2, d = ar3
Now,
if
or if
or if
or if
which is true.
Hence, if a, b, c d are in G.P. then
If a, b, c are in G.P. and x, y are the arithmetic means of a, b and b, c respectively,
then prove that and
Since a, b, c are in G.P.
∴
...(i)
Since x is the A.M. between a and b
∴ ...(ii)
Since y is the A.M. between b and c
∴ ...(iii)
[By using (i)]
∴
Now,
[By using (i)]
∴
If a, b, c are in A.P. and x,y,z are in G.P., show that
Since a, b, c are in A.P.
∴ b - a = c - b 2b = c + a
Here, x, y, z are in G.P.
Let r be the common ratio
∴ y = xr, z = xr2
Now,
[By using (i)]
= 1 . 1 = 1
∴
Let the G.P. be a, ar, ar2, ..........
Given a + ar = -4 ...(i)
Also,
or
(∵
)
Putting r = 2 in (i), we get a + 2a = -4 a = -4/3.
∴ G.P. is
Now, putting r = -2 in (i), we get a + a(-2) = -4 -a = -4 or a = 4.
∴ G.P. is 4, 4(-2), 4(-2)2, .... or 4, -8, 16, ...........
Let a, b be roots of required quadratic equation
Given and
a + b = 16 and ab = 25
S = sum of roots = a + b = 16.
and P = Product of roots = ab = 25
∴ Required quadratic equation is or
Given a, b, c, d are in G.P.
b = ak, c = bk, d = ck or b = ak, c = ak2, d = ak3 ...(1)
Given a, b are roots of equation a + b = 3, ab = p [By using (1)]
a + ak = 3, a. ak = p
a(1 + k) = 3 ...(2)
and ...(3)
Also, c, d, are roots of equation c + d = 12, cd = q
or [By using (1)]
...(4)
and ...(5)
Dividing (4) by (2), we get
...(6)
Consider
Applying componendo and dividendo, we get or (q + p) : (q - p) = 17 : 15.
Let the four numbers in G.P. be
Since their product is 4096.
∴
Since their sum is 85
∴
Put
∴ or
Putting 2t = x, we get
By inspection x = 5 satisfies the equation. So x - 5 is a factor of . Other roots are given by
x is imaginary
∴ x = 5 is the only real value
So, 2t = 5,
When a = 8, r = 2, numbers are
When a = 8, numbers are
or 64, 16, 4, 1
Here,
Therefore,
{The series is 2nd bracket is G.P. with common ratio 2>1}
Insert 5 G.M.'s between
Let be five G.M.'s between
and
∴ are in G.P.
Here, Let r be the common ratio
Now,
When r =
then
When then
Sponsor Area
Consider an identity
Substituting x = 1, 2, 3,........n, we get
............................
.............................
Adding vertically, we get
∴ or
Consider an identity
Substituting
.............................................................
.............................................................
Adding vertically, we get
]
∴ or
Find the sum to n terms of the series:
1 . 2 . 4 + 2. 3 . 7 + 3. 4. 10 +..............
Let Tn be the nth terms of the series
Tn = [nth terms of 1, 2, 3, .......] [ nth terms of 2, 3, 4, ......]
[nth term of 4, 7, 10, ........]
= [ 1 + (n -1) 1] [2 + (n - 1)1] [4 + (n - 1)3]
= (1 + n - 1) (2 + n - 1) (4 + 3n - 3) = n(n + 1) (3n + 1) Tn = 3n3 + 4n2 + n
Let Sn denote the sum of n terms of the series:
∴
=
=
=
The given series is: Let Tn be its nth term
∴
Let denote the sum of n terms
∴ =
=
Find the sum to n terms the series and deduce the sum to infinity.
The given series is:
Let Tn be its nth term
Let be the sum of n terms of the series:
=
DEDUCTION: approach to zero as n approach to infinity
∴
Sum of the series n . 1 + (n - 1). 2 + (n - 2) . 3 + ........ + 1. n
The given series is: n. 1 + (n - 1) . 2 + (n - 2) . 3 + ............. + 1. n
Let Tk denote its kth term
∴ Tk = [kth term of n, n - 1, .....] [kth term of 1, 2, 3, ........]
= [n + (k - 1) (-1)] [1 + (k - 1)1] = (n - k + 1) (k) = nk - k2 + k
Tk = (n + 1)k - k2
Let denote the sum of the series
∴
=
=
Let n + 1, n + 2, .....(n + m) be m consecutive integers
Let denote the sum of the cubes of these integers and
denote the sum of these integers
=
=
=
=
= [1 + 2 + 3 +........ + n + (n + 1) + ..... + (n + m)] - [1 + 2 + ....... + n]
which is an integer because (n + m) (n + m + 1) and n(n + 1) are both even numbers being the product of two consecutive integers. Hence, the sum of cubes of any number of consecutive integers is divisible by the sum of these integers.
Find the sum to n terms of the following series:
52 + 62 + 72 + ....+ 202.
Here, [nth term of 1, 2, 3,......]
[nth term of 2, 3, 4,.....]
[nth term of 3, 4, 5......]
= [1 + (n - 1). 1] [2 + (n - 1). 1] [3 + (n - 1) .1]
=
Let denote the sum of n terms of series, then
=
=
Let
Re-writing it as
Subtracting,
∴
or
∴
or
Let the number of students in the class be n.
Let a be the age of youngest student
∴ a = 8 years
Since the age of the students form an A.P. with common difference 4 months or years.
∴ years
Since the sum of the ages of all the students in the class is 168 years.
∴
n = 16
∵ Number of students cannot be negative. Hence, number of students in the class is 16.
A man deposited Rs. 10,000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
Money deposited = Rs 10,000, Rate of interest = 5%
Interest for one year =
Let a be the amount deposited and d be the interest every year.
∴ a = 10000, d = 500
Amount in the 15th year =
Let denote the total amount after 20 years
= 10 [29500] = Rs. 29500
Let a mm be the distance travelled in the first second. Since the insect covers half the distance covered in the previous second in the succeeding second.
∴ Distances covered in various seconds are
Let r be the common ratio
∴
Let 3 mm distance be covered in n seconds
∴
which is impossible for any value of n.
Hence, the insect would never reach a point 3 mm away from its starting point.
Let the number of sides of the polygon be n. The interior angles from an A.P. Let a be the smallest angle and d be the common difference.
∴
Since the sum of all the interior angles of a polygon with n sides is
∴
n = 9, 16
But if n = 16, then which is impossible.
∵ No angles of polygon exceeds
Hence, n = 9
Original cost of scooter = Rs. 22000
Amount paid in cash = Rs. 4000
Amount to be paid in instalments = Rs. 18000
Let a denote the first instalment
a = 1000 + 10% interest on unpaid amount
= Rs . a = Rs. 2800
2nd instalment =
Similarly, 3rd instalment = Rs. 2600
∴ Various instalments are Rs. 2800, Rs. 2700, Rs. 2600.........
Amount paid in instalments = Rs. 2800 + Rs . 2700 + Rs . 2600 + ......+ Rs. 1100
= Rs .
= Rs. 9[5600 - 1700] = Rs. 9(3900) = Rs. 35100
Actual amount by Hari = Rs. [35100 + 4000] = Rs 39100
Let Rs. a be the value of the certificates purchased by him in the first year and Rs. d be the value of the certificate in excess every year.
∴ d = Rs. 25
Since the value of the certificates purchased by him after 20 years = 7250
∴
2a + 475 = 725
2a = 250
a = Rs. 125
∴ Value of the certificates purchased by him in the first year = Rs. 125 New line Value of the certificates purchased by him in the 13th year = a + (13 - 1)d = 125 + 12(25) = Rs. 425
Let the total number of stones be 2n + 1
∴ There are n stones on each side of the middle stone. Let the man starts collecting stones from the extreme left stone. The distance covered by man to bring the extreme left stone to the middle stone = 10n metres. The distance covered by man to bring the (n - 1)th stone on left to the middle stone = 2 10 (n - 1) metres.
∴ The total distance covered by the man to bring all the stones from left to the middle stone
= 10n + 20(n - 1) + 20(n - 2) + ......... + 20
Similarly, the total distance covered by the man to bring all the stones from the right to the middle stone
= 20 + 20 (n - 1) + ......... + 20
∴ Total distance covered by the man to bring all the stones to the middle stone
= 30n + 40 (n - 1) + 40 (n - 2) +........ + 40 (1)
= 40 [n + (n - 1) + (n - 2) + ....... + 1] - 10n =
But the distance covered is 3 km or 3000 metres.
∴
n = 12
∴ Number of stones = 2(12) + 1 = 25
Here, a = 3,00,000, d = 10,000, n = 20
∴
= 10 (7,90,000) = 79,00,000.
After the end of one year, value of machine = Rs.
= Rs
After the end of two years, value of machine = Rs.
= Rs. 15625
Thus, the value of machines after successive years are......... which form a G.P. with a = Rs. 15625
∴
A square is drawn by joining the mid-points of the sides of a given square. A third square is drawn inside the second square in the same way, and the process continues indefinitely. If a side of the first square is 16 cm., determine the sum of the areas of all the squares.
A class has 30 students. In how many ways can three prizes be awarded so that:
(a) no students get more than one prize?
(b) a student may get any number of prizes?
(a) Number of ways in which the first prize can be awarded = 30 m = 30
Number of eligible students for the second prize = 29
∴ Number of ways in which second prize can be awarded = 29 n = 29
Number of ways in which the third prize can be awarded = 28 P = 28
∴ By fundamental principle of counting, the total number of ways of awarding the three prizes.
=
(b) Number of students = 30
Number of ways in which the first prize may be awarded = 30 m = 30
Number of students eligible for the second prize = 30 (∵ A student may get any number of prizes)
Number of ways in which the second prize can be awarded = 30 n = 30
Similarly, number of ways in which the third prize can be awarded = 30 p = 30
By fundamental principle of counting, the number of ways in which the three prizes can be awarded
=
Event 1: A coin is tossed and the outcomes recorded.
Number of outcomes m = 2
Event 2: The coin is tossed again and the outcomes recorded.
Number of outcomes n = 2
Event 3: The coin is tossed third time and the outcomes recorded.
Number of outcomes p = 2
∴ By fundamental principle of counting, the total number of outcomes recorded
=
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is allowed.
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed.
Number of ways in which place (x) can be filled = 5
m = 5
Number of ways in which place (y) can be filled = 4 (∵ Repetition is not allowed)
n = 4
Number of ways in which place (z) can be filled = 3 (∵ Repetition is not allowed)
p = 3
∴ By fundamental principle of counting, the total number of 3 digit numbers formed
= m x n x p = 5 x 4 x 3 = 60.
Five of the 10-digits (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9) are to be used.
The number must start with 67.
Number of ways of filling box (v) = 1 (∵ Only by 6) m = 1
Number of ways of filling box (u) = 1 (∵ Only by 7) n = 1
Number of ways of filling box (z) = 8 (∵ 6 and 7 are not allowed) p = 8
Number of ways of filling box (y) = 7 (∵ Repetition is not allowed) q = 7
Number of ways of filling box (x) = 6 (∵ Repetition is not allowed) r = 6
∴ Total number of 5-digit telephone numbers formed = m x n x p x q x r = 1 x 1 x 8 x 7 x 6 = 336
Total flags = 5
Number of signals generated, using two flags:
= m x n = 5 x 4 = 20 ...(i)
Or
Number of signals generated, using three flags:
= m x n x p = 5 x 4 x 3 = 60 ...(ii)
Or
Number of signals generated, using 4 flags:
= m x n x p x q = 5 x 4 x 3 x 2 = 120.
Or
Number of signals generated, using all 5 flags:
= m x n x p x q x r = 5 x 4 x 3 x 2 x 1 = 120
∴ Total number of signals generated
= 20 + 60 + 120 +120 = 320.
Digits available are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Total number of digits = 10
Number of digits used = 3
Number of filling box (z) = 9 [∴ If 0 is put in there, it becomes a two digit number] m = 9
Number of ways of filling box (y) = 9 (3 Repetition is not allowed) n = 9
Number of ways of filling box (x) = 8 (3 Repetition is not allowed) p = 8
∴ The number of 3 digit numbers or numbers between 100 and 1000
= m x n x p = 9 x 9 x 8 = 648
Solution not provided.
Ans. 60
Solution not provided.
Ans. 12
Solution not provided.
Ans. 12
How many:
(a) 2-digit even numbers (b) 3-digit even numbers
can be formed using the digits 1, 2, 3, 4, 5 when repetition of digits is not allowed?
Solution not provided.
Ans. (a) 8 (b) 24
Solution not provided.
Ans. (a) 10000 (b) 5040
Solution not provided.
Ans. 6840
Solution not provided.
Ans. 27000
Find the number of 3-digit numbers when
(a) repetition of digit is allowed. (b) repetition of digits is not allowed.
Solution not provided.
Ans. (a) 900 (b) 648
Solution not provided.
Ans. 72
Solution not provided.
Ans. 1024
Solution not provided.
Ans. 8
Solution not provided.
Ans. (i) 180 (ii) 100
Solution not provided.
Ans. 90
3! + 4! = (3 x 2 x 1) + (4 x 3 x 2 x 1) = 6 + 24 = 30
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040
∴
Evaluate: when
(a) n = 6, r = 2, (b) n = 6, r = 4, (c) n = 9, r = 4
(a) n = 6, r = 2
(b) n = 6, r = 4
(c) n = 9, r = 4
Evaluate: when:
(a) n = 6, r = 2, (b) n = 7, r = 7, (c) n = 9, r = 0
(a) n = 6, r = 2
(b) n = 7, r = 7
(∵ 0! = 1)
(c) n = 9, r = 0
Evaluate: when:
(a) n = 9, r = 0, (b) n = 8, r = 1, (c) n = 9, r = 9
(a) n = 9, r = 0
(∵ 0! = 1)
(b) n = 8, r = 1
(c) n = 9, r = 9
(∵ 0! = 1)
If (n+1)! = 30 (n - 1)!, find value of n.
L.H.S. = n! + (n + 1) ! = n! + [n + 1] x n! = n! [1 + n + 1] = n! (n+2)
= [n + 2] x n! = R.H.S.
∴ L.H.S. = R.H.S.
Hence, n! + (n + 1)! = (n + 2) x n!
32! = 32 x 31 x 30 x 29 x ......... x 3 x 2 x 1
The product contains 32 factors in all of which 16 are even factors. Each even number is divisible by 2.
∴ 32! is divisible by 216.
Also, the powers of 2 contained in 32!
25 from 32, 21 from 30, 22 from 28, 21 from 26, 23 from 24, 21 from 22, 22 from 20, 21 from 18,
24 from 16, 21 from 14, 22 from 12, 21 from 10, 23 from 8, 21 from 6, 22 from 4 and 21 from 2.
∴ divides 32!
divides 32!
Hence, the largest value of n for which divides 32! is n = 31.
Prove that 33! is divisible by What is the largest value of n so that 33! is divisible by
?
Solution not provided.
Ans. n = 31
Find r, if
(r - 10) (r - 3) = 0
r = 10, r = 3
But, for
But, [Not valid]
Hence, only value of r = 3
Number of digits available = 9
(1, 2, 3, 4, 5, 6, 7, 8, 9)
Number of digits to be used = 3
(No digit is to repeat)
n = 9, r = 3
Number of permutations =
Hence, the number of 3-digit numbers formed = 504.
How many 4-digit numbers are there if no digit is repeated?
Or
How many numbers are there between 1000 and 9999 so that no digit is repeated?
Number of digits available = 10
(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
Number of digits to be used = 4
Since the numbers are to be of four digits.
∴ 0 cannot be filled in box 1.
Number of permutations for box 1 = 9 ...(i)
Now, we have 9-digits and 3 boxes. n = 9, r = 3.
Number of permutations =
...(ii)
From (i) and (ii), we have total number of permutations, by fundamentals principle of counting
= 9 x 504 = 4536.
The word is 'EQUATION'
Number of letters = 8 (all distinct)
Number of letters to be used = 8 Number of permutations =
= 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320
Additional part:
Fix E in box 1 and N in box 8
Number of permutations for box 1 = ...(i)
Number of permutations for box 8 = ...(ii)
Number of letters left = 6 n = 6
Number of boxes left = 6 r = 6
∴ Number of permutations = ...(iii)
From (i), (ii) and (iii), by fundamental principle of counting, the toal number of words formed
= 1 x 1 x 720 = 720.
Number of letters in word 'Monday' = 6 (all distinct)
Number of vowels = 2 (O and A)
4 letters to be used, n = 6, r = 4
Number of permutations =
Hence, number of words = 360.
Number of letters in word 'Monday' = 6 (all distinct)
Number of vowels = 2 (O and A)
All six letters to be used. n = 6, r = 6
Number of permutations =
Hence, the total number of words formed = 720.
Number of letters in word 'Monday' = 6 (all distinct)
Number of vowels = 2 (O and A)
All six letters to be used but first letter is a vowel.
Number of arrangements for a vowel in box 1 = ...(i)
(Two vowels and one box)
We are left with 5 letters and 5 boxes.
∴ Number of permutations = ...(ii)
∴ From (i) and (ii), by fundamental principle of counting, total words formed
= 2 x 120 = 240
Number of letters in word 'MONDAY' = 6 (all district)
Number of vowels = 2 (O and A)
Number of letters to be used in 4 and first letter is a vowel,
Number of arrangements for box 1 = (i)
(Two vowels for 1 box)
We now have 5 letters left and 3 boxes for them. n = 5, r = 3
∴ The number of permutations ...(ii)
Hence, from (i) and (ii), the total number of permutations i.e., the number of words formed:
= 2 x 60 = 120.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Number of letters = (all distinct)
Number of vowels = 5 (e, i, o, u, a)
Number of consonants = 3 (q, t, n)
Step I: Tie the vowels together Number of permutations =
Step II: Tie the consonants together. Number of permutations =
Step III: Mix the bundles and any remaining letter to give 1 + 1 + 0 = 2 letters.
Number of permutations =
Hence, the number of permutations, using, fundamental principle of counting
= 120 x 6 x 2 = 1440
How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that no two consonants are together.
Number of letters = (all distinct)
Number of vowels = 5 (e, i, o, u, a)
Number of consonants = 3 (q, t, n)
Arrange the vowels in a row by leaving one place between every two vowels.
Number of permutations of arrranging vowels ...(i)
V
V
V
V
V
1 2 3 4 5 6
Number of places for the consonants, so that no two of them are together = 6
Number of consonants = 3 n = 6, r = 3
Number of permutations of arranging consonants:
...(ii)
Hence, using (i) and (ii), the number of words formed so that no two consonants are together, using fundamental principle of counting
= 120 x 120 = 14400
No two boys are together.
Step I: Arrange 6 girls in a row by leaving one seat between every two girls.
Number of permutations of arranging girls = 6! = 720 ...(i)
G
G
G
G
G
G
1 2 3 4 5 6 7
Step II: There are 7 places for the boys to sit in order that no two of them are together.
Number of boys = 4
n = 7, r = 4
∴ Number of permutations of arranging boys = ...(ii)
Hence, from (i) and (ii),
The number of ways in which they may be seated so that no two boys are together
= 720 x 840 = 604800
Number of letters in word 'ORIENTAL' = 8 (all distinct)
Number of letters to be used = 8
Step I: A and E are to occupy odd places marked X
Number of letters = 2 (A and E)
Number of boxes = 4
n = 4, r = 2
Number of permutations of arranging A and E = ...(i)
Step II: After A and E are fixed, there will be 6 letters left and 6 boxes for them.
(A and E use up two boxes)
Number of permutations of arrranging the remaining letters =
...(ii)
Hence, from (i) and (ii), the total number of words formed = 12 x 720 = 8640.
In how many ways can 5 girls be seated in a row so that two girls Ridhi and Sanya are:
(a) always together (b) never together
Number of girls = 5
(a) Consider Ridhi and Sanya together as one girl.
Now, number of girls becomes 4
Number of permutation of arranging these 4 girls =
But, the two girls Ridhi and Sanya can be arranged in ways or 2! ways or 2 ways.
By fundamental principle of counting, the required number of permutations = 24 x 2 = 48
(b) Total number of permutations =
Number of permutations in wich Ridhi and Sanya are never together:
= 120 - 48 = 72
Number of digits = 6 (all different)
Number of digits used = 6
(a) Digits are not repeated n = 6, r = 6 Number of permutations =
Hence, the numbers formed = 720.
(b) The digits may be repeated.
Number of arrangements =
Hence, numbers formed = 46656.
In part (a), the numbers are to be odd.
Fix box 6 for an odd number.
Number of odd digits = 3 (5, 7, 9)
Number of boxes = 1 n = 3, r = 1
Number of permutations =
The remaining 5-digits are to be arranged in 5 boxes.
Number of permutations =
∴ Total numbers formed = 3 x 120 = 360.
Number of digit available = 5 (all distinct)
Number of digits to be used = 5
Since the number is not divisible by 5. 5 cannot occupy box 5.
Number of permutations for box ...(i)
Now, we have 4-digits and 4 boxes.
Number of permutations = ... (ii)
From (i) and (ii), the total number of permutations = 4 x 24 = 96
Hence, the number of numbers formed, that are not divisible by 5 = 96.
The number of ways in which first prize may be given = 10.
Number of ways in which second prize may be given = 10.
Number of ways in which third prize may be given = 10.
∴ By fundamental principle of counting, the total number of permutations
= 10 x 10 x 10 = 1000
ALTERNATIVELY : n = 10, r = 3
Number of permutations (for repeat case) = nr = 103 = 1000.
Number of digits = 10 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
Number of digits used = 4
For a 4-digit number, zero may not occupy box 1.
Number of permutations for box 1 =
Number of ways filling boxes 2, 3, 4 =
Total number of 4-digit numbers = 9 x 1000 = 9000.
Number of digits = 10 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
Number of digits used = 4
The digits are not to be repeated.
Number of permutations for box 1 = 9 (zero cannot be placed in box 1)
We have 9-digits left (no repetition is allowed) and three boxes. Number of permutations
By fundamental principle of counting, the number of permutations, when a digit is not repeated
= 9 x 504 = 4536.
Number of digits = 10 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
Number of digits used = 4
Number of permutations in which at least one digit is reveresed = x - y ...(i)
where x = number of permutations when any digit may be repeated any number of times,
y = number of permutations when no digit is repeated.
From part (a), x = 9000
From part (b), y = 4536.
Hence, the number of 4-digit numbers in which at least one digit is repeated.
= 9000 - 4536 = 4464
(b) there is no restriction as to the number of prizes that a boy may get.
(c) no boy sets all prizes.
(a) n = 6, r = 3
Number of permutations =
Hence, the number of ways, in which the three prizes can be awarded = 120.
(b) n = 6, r = 3
Number of permutations =
Hence, the number of ways in which 3 prizes can be awarded = 216.
(c) Number of ways in which one boy gets all prizes = number of boys = n = 6.
There are 8 students appearing for an examination, of which 3 appear in mathematics paper, and 5 in other different subjects. In how many ways can they be seated if
(a) all the students appearing for mathematics paper are together.
(b) the students appearing for mathematics paper are not all together.
(c) no two students appearing in mathematics paper are seated together ?
Total number of books = n + n = 2n
Let the places on the shelf for these books are: 1, 2, 3, 4, 5, 6, ..........(2n - 3), (2n - 2), (2n - 1), 2n.
M C M C M C M C M C
Case (i)
Mathematics books are arranged on odd number places and chemistry books are placed on even number places.
Number of arrangements =
Or
Case (ii)
Chemistry books are placed an odd number places and mathematics books on even number places.
Number of arrangements =
Hence, the total number of arrangements on the shelf where mathematics books and chemistry books are placed alternatively.
Solution not provided.
In how many ways can the letters of the word ‘PENTAGON’ be permuted, using all letters ?
(a) How many of them begin with P and end with N ?
(b) How many of them have the three vowels together ?
Solution not provided.
Ans.(a) 720
(b) 4320
Solution not provided.
Ans.(a) 720
(b) 120
Solution not provided.
Ans. 1440
Find the number of words formed by arranging the letters of the word ‘DAUGHTER’ so that
(a) not all vowels are together (b) no two vowels are together.
Solution not provided.
Ans. (a) 36000 (b) 14400
Solution not provided.
Ans. 120; 60
(a) the girls are together? (b) all the three girls are not together?
(c) no two girls are together?
Solution not provided.
Ans. (a) 4320, (b) 36000, (c) 14400
Solution not provided.
Ans. 60 ; 24
Solution not provided.
Ans. 56
(a) How many 4-digit numbers are there without repetition of digits?
(b) How many 4-digits are there, if there is no restriction on repetition of digits?
(c) How many 4-digit numbers are there with at least one number repeated?
Solution not provided.
Ans. (a) 4536
(b) 9000
(c) 4464
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Solution not provided.
Ans. 60
Solution not provided.
Ans. 11880
Solution not provided.
Ans. 60
Solution not provided.
Ans. 336
Solution not provided.
Ans. 32
Solution not provided.
Ans. 1024
(a) no student gets more than one prize?
(b) any student may get any number of prize (s)?
(c) no student may get all prizes?
Solution not provided.
Ans. (a) 336
(b) 512
(c) 504
Solution not provided.
Ans. 81
Solution not provided.
Ans. 1036800
Find the number of ways in which 6 objects out of 8 can be arranged so that two particular objects are :
(a) included (b) excluded (Hint : (a) 6P2 x 6P4 (b) 6P6)
Solution not provided.
Ans. (a) 10800, (b) 720
(i) first letter is a vowel (ii) no two vowels are together
(iii) relative position of vowels and consonants remains unchanged?
Solution not provided.
Ans. (i) 2160, (ii) 1440, (iii) 144
(i) How many different numbers of 6-digits can be formed with the numbers 3, 5, 7, 8, 2, 6?
(ii) How many of them are divisible by 5?
Solution not provided.
Ans. (i) 720
(ii) 120
Solution not provided.
Ans. 12
Solution not provided.
Ans. 72
Solution not provided.
Ans. 8 x 9!
(ii) no two vowels are together ?
(iii) vowels may occupy odd places ?
Solution not provided.
Ans. (i) 720
(ii) 1440
(iii) 576
(i) exactly three flags can be used for a signal?
(ii) at most three flags are to be used for a signal?
(iii) at least three flags are to be used for a signal?
Solution not provided.
Ans. (i) 60, (ii) 85, (iii) 300
Total number of balls =
n = 9
Number of red balls = 4 p = 4
Number of yellow balls = 3
q = 3
Number of green balls = 2 r = 2
∴ The number of permutations of balls:
= 9 x 4 x 7 x 5 = 1260.
The letter of the word MISSISSIPPI are:
M → 1
I → 4
S → 4
P → 2
Total → 11 n = 11, p = 4, q = 4, r = 2
∴ The number of permutations =
= 11 x 10 x 9 x 7 x 5 = 34650.
Tie the 4 1's together.
Number of permutations =
(∵ 4 I's are similar)
= ...(i)
Mix the bundle of I's with IM + 4S + 2P to give a total of (1 + 4 + 2) + 1 = 8
∴ The number of permutations =
Hence, the number of permutations, when the four I's are together = 840 x 1 = 840
The number of permutations when the four I's are not all together
(Total permutations) - (permutations when I's are together) = 34650 - 840 = 33810.
Number of letters:
A → 3, S → 4
I → 2, N → 2
T → 1, O → 1
Total letters 13.
Tie the 4 S's.
Number of permutations =
Mix with remaining to give:
[3 (A) + 2 (I) + 2 (N) + 1 (T) + 1 (O)] + 1 = 10.
Number of arrangements =
Total number of arrangements = 1 x 151200 = 151200.
Number of letters:
A → 3, S → 4
I → 2, N → 2
T → 1, O → 1
Total letters 13.
Tie the four S's.
Number of arrangements = ...(i)
Mix with remaining to give:
[3 (A) + 2 (I) + 2 (N) + 1 (T) + 1 (O)] + 1 = 10
Fix O at first place.
Number of permutations = ...(ii)
Fix T at the last place.
Number of permutations = ...(iii)
Now, we have
3 (A) + 2 (I) + 2 (N) + 1 (four S's tied) = 8 letters
Number of arrangements =
Hence, by fundamental principle of counting, the total number of permutations
= 1 x 1 x 1 x 1680 = 1680.
How many different words, with or without meaning can be formed, by using the letters of the word ‘HARYANA’? Also, find as to:
(a) how many of these begin with H and end with N?
(b) in how many of these H and N are together?
Total letters in 'HARYANA' = 7
H → 1, A → 3, R → 1, Y → 1, N → 1. n = 7, p = 3
∴ Number of arrangements =
(a) The word begins with H and ends with N.
Fix the first place with H.
The number of permutations = ...(i)
Fix the last place with N.
The number of permutations = ...(ii)
Now, we are left with 5 places and 3(A) + 1 (R) + 1 (Y) Number of arrangements =
...(iii)
∴ By fundamental principle of counting, the number of arrangements
= 1 x 1 x 20 = 20.
(b) When H and N are together.
Tie the two.
Number of arrangements = ...(i)
Mix the tied bundle with the remaining.
[3 (A) + 1 (R) + 1(Y)] + 1 = 6 letters
Number of permutations = ...(ii)
∴ From (i) and (ii), applying fundamental principle of counting, we get the number of arrangements when H and N are together
= 2 x 120 = 240.
Total letters in word 'AGAIN' (A → 2, G → 1, I → 1, N → 1) = 5
Number of words that begin with A =
(Fix A at first place, so that the remaining 4 letters are different)
Number of words with begin with G =
Number of words that begin with I =
Thus, so far we have formed 24 + 12 + 12 = 48 words.
The 49th word will begin with N and it is NAAGI and the 50th word is NAAIG.
Numbers of letters in 'RANDOM' = (R → 1, A → 1, N → 1, D → 1, O → 1, M → 1) = 6
Number of words that begin with A =
Number of words that begin with D =
Number of words that begin with M =
Number of words that begin with O =
Number of words that begin with N =
So far, we have formed 120 + 120 + 120 + 120 + 120 = 600 words.
Now words starting with R.
Number of words starting with RAD = 1 x 1 x 1 x 3! = 6.
Number of words starting with RAM = 1 x 1 x 1 x 3! = 6.
Total words = 600 + 12 = 612.
613th word is RANDMO
614th word is RANDOM
∴ Rank of the word random in a dictionary is 614.
How many different words can be formed by using the letters of the word ‘ALLAHABAD?
(a) In how many of these do the vowels occupy even positions.
(b) In how many of these, the two L’s do not come together?
Number of letters in word 'ALLAHABAD' = (A → 4, L → 2, H → 1, B → 1, D → 1) = 9
Number of arrangements =
(a) There are only four A's as vowels.
They can occupy even places (2, 4, 6, 8) in ways
∴ Number of ways in which vowels occupying even places = 1
We are left with 5 places and letters (L → 2, H → 1, B → 1, D → 1).
Number of permutations =
Hence, total number of arrangements in which A's occupy even places
= 1 x 60 = 60.
(b) We first find the number of arrangements in which two L's are not together:
Number of arrangements in which two L's are together
Hence, the number of arrangements in which the two L's are not together
= (Total arrangements) - (the number of arrangements in which the two L's are together)
= 7560 - 1680 = 5880.
In how many ways can 8 students be seated
(a) in a straight line (b) in a circle?
n = 8
(a) Number of arrangements in a row =
(b) Number of arrangements in circular formation = (n - 1)! = 7! = 5040.
Tie the two
Number of arrangements =
Mix with remaining
Now, we have 3 + 1 = 4 persons to sit around a round table.
The number of permutations = (4 - 1)! = 3! = 6
Hence, the total number of arrangments when the two persons sit together
= 2 x 6 = 12
Find the number of permutations of 6 students sitting around a round table.
(a) In how many of these arrangements are three of the students sit together ?
(b) In how many of the arrangements, three of the students do not sit all together ?
Number of arrangements for 6 students to sit around a round table = (6 - 1)! = 5! = 120.
(a) Tie the three students.
Number of arrangements =
Mix with the remaining to give a toal of 3 + 1 = 4
Seat them around a round table.
∴ The number of permutations = (4 - 1)! = 3! = 6.
Hence, the number of arrangements in which the 3 students sit together = 6 x 6 = 36
[By using (i) and (ii)]
(b) The number of arrangements in which the 3 students do not sit all together = 120 - 26 = 84.
Let the boys sit around the round table.
Number of arrangements = (4 - 1)! =3! = 6 ...(i)
There are 4 seats for the girls and there are 3 girls.
Number of arrangements = ...(ii)
∴ Total number of permutations when no two girls are together = 6 x 24 = 144
(a) In how many ways can a girl make a garland with 8 flowers?
(b) In how many ways can a girl make a necklace with 4 red, 3 green and 2 blue beads?
(a) Number of flowers = 8
n = 8
Number of ways in which the garland can be made
(b) Number of beads = (4 red + 3 green + 2 blue) = 9
Number of ways in which the necklace can be made
=
=
= 70.
(a) INDIA (b) ARRANGE (c) EXERCISES
(d) SERIES (e) INTERMEDIATE
Solution not provided.
Ans. (a) 60 (b) 1260 (c) 30240 (d) 180 (e) 19958400
Find the number of arrangements of the letters of word ‘INDEPENDENCE’. How many of these words:
(a) start with P? (b) the vowels always occur together?
(c) the vowels never occur together?
(d) begin with I and end with P?
Solution not provided.
Ans. 1663200 (a) 138600 (b) 16800 (c) 1646400 (d) 12600
How many words can be formed using the letters of word ‘PARALLEL’ so that the L’s
(a) come together (b) do not come together ?
Solution not provided.
Ans. (a) 360 (b) 200
Solution not provided.
Ans. 60480
Solution not provided.
Ans. 14
Solution not provided.
Ans. 900
Solution not provided.
Ans. (a) 6 (b) 12
(a) How many of them begin with C?
(b) How many begin with T?
(c) In how many of these vowels occur together?
Solution not provided.
Ans. 4989600 (a) 453600 (b) 907200 (c) 120960
Solution not provided.
Ans. 907200
Solution not provided.
Ans. 144
Solution not provided.
Ans. (a) 12 (b) 12
Solution not provided.
Ans. 309
Solution not provided.
Ans. 60
(a) In how many ways can 6 white beads and 5 red beads be arranged in a line on a table?
(b) In how many ways can 6 white and 5 red beads form a necklace?
Solution not provided.
Ans. (a) 462 (b) 21
Solution not provided.
Ans. 18
Solution not provided.
Ans. 48
Solution not provided.
Ans. (i) 180; (ii) 150
Solution not provided.
Ans. 13! - 2! x 12!
Number of boys = 5
Number of boys to be selected = 3 Number of combinations (selections) =
...(i)
Number of girls = 4
Number of girls to be selected = 3 Number of selections =
...(ii)
∴ From (i) and (ii), by fundamental principle of counting, the number of selections
= 10 x 4 = 40
Hence, the number of ways in which 3 boys and 3 girls out of 5 boys and 4 girls can be selected = 40
Number of black balls = 5
Number of black balls to be selected = 2 Number of selections =
...(i)
Number of red balls = 6
Number of red balls to be selected = 3 Number of selections =
Hence, from (i) and (ii), by fundamental principle of counting, the required number of selections
= 10 x 20 = 200.
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Number of players = 17 (5 bowlers + 12 others)
Number of players to be selected = 11 (4 bowlers + 7 others)
Number of selections for 4 bowlers out of 5 = ...(i)
Number of selections for the remaining 7 players out of 12
=
= ...(ii)
Hence, from (i) and (ii), by fundamental principle of counting, the number of ways in which the team can be selected = 5 x 792 = 3960
Number of consonants = 21
Number of vowels = 5
A word contains two vowels and two consonants will be formed as
(i) select 2 vowels out of 5
(ii) select 2 consonants out of 21
(iii) arrange 2 + 2 = 4 letters to obtain different words
(iv) use fundamental principle of counting.
(i) Number of ways in which 2 vowels can be selected out of 5 =
(ii) Number of ways in which 2 consonants can be selected out of 21 =
(iii) Number of permutations (to form different words) of the four letters = 4!
Hence, by fundamental principle of counting, the number of words formed.
= 10 x 210 x 24 = 50400
The word is 'INVOLUTE'
Number of consonants = 4
Number of vowels = 4.
The words formed should contain 3 vowels and 2 consonants.
The problems becomes:
(i) Select 3 vowels out of 4. Number of selections =
(ii) Select two consonants out of 4. Number of selections =
(iii) Arrange the five letters (3 vowels + 2 consonants) to form words.
Number of permutations = 5!
(iv) Apply fundamental principle of counting:
Number of words formed =
=
= 4 x 6 x 120 = 2880
Hence, the number of words formed = 2880
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done, when the committee consists of :
(i) exactly 3 girls? (ii) at least three girls? (iii) at most 3 girls?
Number of boys = 9
Number of girls = 4
The committee is to consist of 7
(i) The committee consists of exactly 3 girls Number of boys = 7 - 3 = 4
∴ The number of selections =
=
(ii) The committee consists of at least 3 girls. The committee consists of 3 girls + 4 boys or 4 girls + 3 boys
(Since, the number of girl is 4, so there is no further option)
When committee consists of 3 girls + 4 boys
Number of selections =
When committee consists of 4 girls + 3 boys
Number of selections =
Hence, the number of selections for the committee = 504 + 84 = 588.
(iii) The committee consists of at most 3 girls.
This gives us the following options:
No girl + 7 boys
Number of selections =
Or
1 girl + 6 boys
Number of selections =
Or
2 girls + 5 boys
Number of selections =
Or
3 girls + 4 boys
Number of selections =
Hence, the total number of selections = 36 + 336 + 756 + 504 = 1632.
Total number of boys and girls in the group = 4 girls + 7 boys = 11
Number of boys and girls in the team = 5
(i) The team consists of no girl, The team consists of 0 girl + 5 boys
∴ Number of selections =
Hence, the number of teams formed = 21
(ii) The team consists of at least 1 boy and 1 girl.
Options are:
The team consists of 1 girl + 4 boys
Number of selections =
= 4 x 7 x 5 = 140.
Or
The team consists of 2 girls + 3 boys Number of selections =
=
Or
The team consists of 3 girls + 2 boys Number of selections =
=
Or
The team consists of 4 girls + 1 boy. Number of selections =
Hence, the total number of teams that can be formed = 140 + 210 + 84 + 7 = 441
(iii) The team consists of at least 3 girls
Options are:
The team consists of 3 girls + 2 boys. Number of selections =
Or
The team consists of 4 girls + 1 boy. Number of selections =
Hence, the total number of teams that can be formed = 84 + 7 = 91
Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Total cards = 52
Number of cards to be selected = 5
Type of cards requried = 1 king + 4 others
We have 4 kings + 48 others. Number of selections =
=
= 778320
What is the number of ways of choosing 4 cards from a pack of 52 cards? In how many of these :
(i) four cards are of the same suit?
(ii) four cards belongs to different suit?
(iii) are face cards?
(iv) two are red cards and two are black cards?
(v) cards are of the same colour?
Number of cards in the deck = 52
Number of cards to be selected = 4 Number of selections =
= 270725
(i) The four cards are of the same suit.
Options are:
They are all club cards Number of selections =
Or
They are all spade cards. Number of selections =
Or
They are all diamond cards. Number of selections =
Or
They are all heart cards. Number of selections =
Hence, the total number of selections =
=
(ii) They are one from each suit. 1 card is club and 1 card is spade and 1 card is diamond and 1 card is heart.
Total number of combinations =
= 13 x 13 x 13 x 13
= 28561.
(iii) The four cards are face cards.
Number of face cards = 4 kings + 4 queens + 4 jacks = 12
Number of other cards = 52 - 12 = 40
4 out of 12 face cards are to be selected
Number of selections =
(iv) There are 26 black cards and 26 red cards.
Two are to be selected from black cards and two from the red cards. Number of selections =
=
= 105625
Hence, the number of selections = 105625
(v) The four cards are of the same colour. All four are either red or all four are black.
Number of selections =
=
=
= 29900.
How many chords can be drawn through 21 points on a circle?
Since 21 points lie on the circumference of a circle.
∴ All the 21 points are distinct and no three of them are collinear. Now, we know that one and only one line can be drawn through 2 distinct points.
∴ Numbers of straight lines formed by 21 points by taking 2 at a time
Hence, the number of chords = 210.
How many straight lines can be formed with 7 points, 3 of which are colinear.
Total number of points = 7.
Number of straight lines formed by these 7 points (no three of which are collinear) by taking 2 at a time
=
It is a given that three points are collinear. If these three points are considered to be distinct, number of straight lines formed by these 3 points by taking 2 at a time =
Instead of these three lines, the 3 collinear points will give us only 1 line.
∴ Required number of straight lines =
Hence, the number of straight lines formed with 7 points, 3 of which are collinear = 19
Number of points = 9
Since 9 points lie on the circumference of a circle.
∴ All the 9 points are distinct and no three of them are collinear.
Now, one and only one triangle can be drawn through 3 distinct and non-collinear points.
∴ Number of triangles formed by 9 points by taking 3 at a time
=
Number of points = 12
Number of traingles formed by these 12 points (no three of which are collinear) by taking 3 at a time
=
It is given that 5 points are collinear.
If these 5 points are considered to be distinct such that no three of them are collinear, then the number of triangles formed by these 5 points by taking 3 at a time
∴ Required number of triangles =
If Find n and r.
Total number of questions:
Part I = 5
Part II = 7
Number of questions to be attempted = 8
Minimum 3 from each part
Options are:
3 form part I, 5 from part II,
Number of selections =
= 210. ...(i)
Or
4 from part I, 4 from part II,
Number of selections =
= 175 ...(ii)
Or
5 from part I and 3 from part II.
Number of selections = ...(iii)
Hence, the total number of selections = 210 + 175 + 35 = 420.
Total number of students = 25
The number of students to go for excursion = 10
Options are:
All 3 of them join the party and 7 others join the party from remaining 22 students. Number of selections =
...(i)
All there do not join the party and 10 join the party from remaining 22 students. Number of selections =
...(ii)
Hence, the total number of ways of selecting 10 out of 25 for an excursion party
=
= 170544 + 646646 = 817190.
Solution not provided.
Ans. 11
(a) In how many of these selections, two particular books are always selected?
(b) In how many of these two books are always excluded?
Solution not provided.
Ans. 210 (a) 28 (b) 70
Solution not provided.
Ans. 15
Solution not provided.
Ans. 778320
Solution not provided.
Ans. 425
Solution not provided.
Ans. 2000
A bag contains 6 balls each of red, blue, green and yellow colours. In how many ways can 4 balls be drawn from the bag?
(a) In how many of the selections, the balls drawn are of the same colour?
(b) In how many of the selections the balls drawn are all of different colours?
Solution not provided.
Ans. 10626 (a) 60 (b) 1296
There are 14 points in a plane, 7 of which are collinear. How many
(a) straight lines (b) triangles can be formed by these points?
Solution not provided.
Ans. (a) 71 (b) 329
Solution not provided.
Ans. 7200
Solution not provided.
Ans. 1596000
Solution not provided.
Ans. (i) 21 (ii) 441 (iii) 91
Solution not provided.
Ans. 105
Solution not provided.
Ans. (i) 420 (ii) 469 (iii) 2954
Solution not provided.
Ans. 15600
How many 3-digit numbers can be formed without using the digit 0, 2, 3, 4, 5, 6 when:
(a) repetition of digits is not allowed ?
(b) repetition of digits is allowed ?
Solution not provided.
Ans. (a) 24 (b) 64
Solution not provided.
Ans. 128
Convert the following to factorials:
(n + 1) (n + 2) (n + 3) ........ (2n)
Solution not provided.
Ans.
What is the largest value of n for which 34! is divisible by ?
Solution not provided.
Ans. 32
Solution not provided.
Ans. 24
Solution not provided.
Ans. 210
Solution not provided.
Ans. 240
Solution not provided.
Ans. 4320
Solution not provided.
Ans. (a) 120 (b) 24
Solution not provided.
Ans. 12
How many words can be formed by using the letters of the word ‘DAUGHTER’ when :
(a) the vowels come together?
(b) the vowels never come together?
Solution not provided.
Ans. (a) 4320 (b) 36000
In how many ways can 5 children be arranged so that two of them, Ram and Rattan
(a) are always together?
(b) are never together?
Solution not provided.
Ans. (a) 48 (b) 72
5 boys and 4 girls are to be arranged in a row so that no two girls are together. In howmany can this be done?
Solution not provided.
Ans. 43200
In how many ways can the letters of the word ‘PERMUTATIONS’ be arranged if the:
(i) words start with P and end with S?
(ii) vowels are all together ?
(iii) there are always 4 letters between P and S?
Solution not provided.
Ans. (i)
(ii)
(iii)
How many words can be formed by using the letters of word ‘VOWELS’ if
(a) there is no restriction on letters?
(b) each word begins with E?
(c) each word begins with E and ends with L?
(d) the vowels are together?
(e) all consonants come together?
Solution not provided.
Ans. (a) 720 (b) 120 (c) 24 (d) 240 (e) 144
In how many ways can a football team of 11 players be selected from 16 players if:(a) two players are always included?
(b) two players are always excluded?
Solution not provided.
Ans. (a) 2002 (b) 364
Solution not provided.
Ans. 210
(a) diagonals (b) triangles formed in a decagon.
Solution not provided.
Ans. (a) 35 (b) 120
Solution not provided.
Ans. (a) 224 (b) 672
Number of ways in which he can enter = 2 m = 2
Number of ways in which he can leave the room = 3 n = 3
By fundamental principle of counting, the number of ways in which he can enter the room from one side and get out from the outer side = m x n = 2 x 3 = 6
If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is:
46th
59th
52nd
58th
D.
58th
Clearly, number of words start with A = 4!/ 2! = 12
Number of words start with L = 4! = 24
A number of words start with M = 4!/2! = 12
Number of words start with SA = 3!/2! = 3
Number of words starts with SL = 3! = 6
Note that, next word will be 'SMALL'
Hence, the position of word 'SMALL' is 58th.
If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:
8/5
4/3
1
7/4
B.
4/3
Let a be the first term and d be a common difference. Then, we have a+d, a+4d, a+8d in GP,
ie. (a +4d)2 = (a+d)(a+8d)
⇒ a2 +16d2 +8ad = a2 +8ad +ad +8d2
⇒ 8d2 = ad
8d =a [∴ d≠0]
Now, common ratio,
If the sum of the first ten terms of the series,is 16/5 m, the m is equal to
102
101
100
99
B.
101
Let S10 be the sum of first ten terms of the series. Then, we have
If m is the AMN of two distinct real numbers l and n (l,n>1) and G1, G2, and G3 are three geometric means between l and n, then equals
4l2 mn
4lm2n
4 lmn2
4l2m2n2
B.
4lm2n
Given,
m is the AM of l and n
l +n = 2m
and G1, G2, G3, n are in GP
Let r be the common ratio of this GP
G1 = lr
G2 =lr2
G3= lr3
n = lr4
Three positive numbers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then, the common ratio of the GP is
D.
Let a, ar, ar2 be in GP (r>1)
On multiplying middle term by 2, a 2ar are in AP.
⇒ 4ar = a+ ar2
⇒ r2-4r +1 = 0
The sum of first 20 terms of the sequence 0.7,0.77,0.777...... is
C.
Let S = 0.7 + 0.77 +0.777 + .... upto 20 terms
If x, y, z are in A.P. and tan−1 x, tan−1 y and tan−1 z are also in A.P., then
x= y= z
2x =3y = 6z
6x = 3y= 2z
6x = 4y = 3z
A.
x= y= z
If x, y, z are in A.P.
2y = x + z and
tan−1 x, tan−1 y, tan−1 z are in A.P.
2 tan−1 y = tan−1 x + tan−1
z ⇒ x = y = z.
If 100 times the 100th term of an AP with non zero common difference equals the 50 times its 50th term, then the 150th term of this AP is
–150
150
times its 50th term
0
D.
0
The 150 th term of this AP
Let a be the first term and d be the common difference of the given AP, then
T100 = a+ (100-1)d = a + 99d
T50 = a +(50-1)d = a +49 d
T150 = a + (150-1) d = a +149 d
Now, according to the question,
100 x T100 = 50 x T50
⇒ 100 (a +99d) = 50(a +49d)
2(a +99d) = (a+ 49d)
2a +198 d =a +49d
a +149d = 0
Let x1, x2, ......, xn be n observations, and let be their arithmetic mean and σ2 be their variance.
Statement 1: Variance of 2x1, 2x2, ......, 2xn is 4 σ2.
Statement 2: Arithmetic mean of 2x1, 2x2, ......, 2xn is 4.
Statement 1 is false, statement 2 is true
Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1
Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1
Statement 1 is true, statement 2 is false
D.
Statement 1 is true, statement 2 is false
A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after
18 Months
19 Months
20 Months
21 Months
D.
21 Months
a = Rs. 200
d = Rs. 40
savings in first two months = Rs. 400
remained savings = 200 + 240 + 280 + ..... upto n terms =
n/2[400 + (n -1)40] = 11040 - 400
200n + 20n2 - 20n = 10640
20n2 + 180 n - 10640 = 0
n2 + 9n - 532 = 0
(n + 28) (n - 19) = 0
n = 19
∴ no. of months = 19 + 2 = 21
A person is to count 4500 currency notes. Let a denote the number of notes he counts in the nth minute. If a1 = a2 = ... = a10 = 150 and a10, a11, ...are in an AP with common difference -2, then the time taken by him to count all notes is
24 min
34 min
125 min
135 min
B.
34 min
Let the first term of an AP be a and common difference be d and number of terms be n, then
tn = a + (n-1)d and Sn = n/2 [ 2a + (n-1)d]
Number of notes that the person counts in 10 min = 10 x 150 = 1500
Since a10, a11, a12, .... are in AP in with common difference -2
Let n be the time has taken to count remaining 3000 notes than
n/2[2 x 148 + (n-1) x -2] = 3000
⇒ n2-149n +3000 = 0
⇒ (n-24)n-125) = 0
n = 24, 125
Then, the total time taken by the person to count all notes = 10 +24 = 34 min
n = 125 is discarded as putting n = 125
an = 148 + (125-1)(-2)
= 148 - 124 x 2 = 148-248 = -100
⇒ Number of notes cannot be negative.
Statement-1: S3 = 55 × 29.
Statement-2: S1 = 90 × 28 and S2 = 10 × 28.
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is false.
Statement-1 is false, Statement-2 is true.
C.
Statement-1 is true, Statement-2 is false.
For any three positive real numbers a, b and c, (25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c), Then
b , c and a are in G.P
b, c and a are in A.P
a, b and c are in A.P
a, b and c are in G.P
B.
b, c and a are in A.P
225a2 + 9b2 + 25c2 – 75ac – 45ab – 15bc = 0
(15a)2 + (3b)2 + (5c)2 – (15a)(3b) – (3b)(5c) – (15a) (5c) = 0
1/2[(15a – 3b)2 + (3b – 5c)2 + (5c – 15a)2] = 0
15a = 3b , 3b = 5c , 5c = 15a
5a = b , 3b = 5c , c = 3a
a/1 = b/5 = c/3
a = λ, b = 5λ, c = 3λ
a, c, b are in AP
How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?
8 . 6C4 . 7C4
6 . 7 . 8C4
6 . 8 . 7C4
7 . 6C4 . 8C4
D.
7 . 6C4 . 8C4
Other than S, seven letters M, I, I, I, P, P, I can be arranged in 7!/2! 4!=7 . 5 . 3.
Now four S can be placed in 8 spaces in 8 C4 ways. Desired number of ways = 7 . 5 . 3 . 8C4 = 7 . 6C4 . 8C4.
How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical order?
120
480
360
240
C.
360
A total number of ways in which all letters can be arranged in alphabetical order = 6! There are two vowels in the word GARDEN. A total number of ways in which these two vowels can be arranged = 2!
∴ Total number of required ways
∴ Total number of required ways
The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is
5
38
21
D.
21
The required number of ways
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