NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Maths
Chapter : 9
Chapter Name : Sequences And Series

Exercise 9.1

Q1 Write the rst ve terms of the sequences whose n th
term is a n = n(n + 2) .

Answer. a = n(n + 2)
n

Substituting n = 1, 2, 3, 4, and 5, we obtain
a1 = 1(1 + 2) = 3

a2 = 2(2 + 2) = 8

a3 = 3(3 + 2) = 15

a4 = 4(4 + 2) = 24

a5 = 5(5 + 2) = 35

Therefore, the required terms are 3, 8, 15, 24, and 35.

Page : 180 , Block Name : Exercise 9.1

Q2 Write the rst ve terms of the sequences whose n term is a .
th n
n =
n+1

Answer. a
n
n =
n+1

Substituting n = 1, 2, 3, 4, 5, we obtain
1 1 2 2 3 3 4 4 5 5
a1 = = , a2 = = , a3 = = , a4 = = , a5 = =
1+1 2 2+1 3 3+1 4 4+1 5 5+1 6

Therefore, the required terms are
1 2 3 4 5
, , , , and
2 3 4 5 6

Page : 180 , Block Name : Exercise 9.1

Q3 Write the rst ve terms of the sequences whose n term is a
th n
n = 2

Answer. a = 2
n
n

Substituting n = 1, 2, 3, 4, 5, we obtain
1
a1 = 2 = 2

2
a2 = 2 = 4

3
a3 = 2 = 8

4
a4 = 2 = 16

5
a5 = 2 = 32

Therefore, the required terms are 2, 4, 8, 16, and 32.

Page : 180 , Block Name : Exercise 9.1

Q4 Write the rst ve terms of the sequences whose n term is a
th 2n−3
n =
6

Answer. Substituting n = 1, 2, 3, 4, 5, we obtain
2 × 1 − 3 −1
a1 = =
6 6

2 × 2 − 3 1
a2 = =
6 6

2 × 3 − 3 3 1
a3 = = =
6 6 2

2 × 4 − 3 5
a4 = =
6 6

2 × 5 − 3 7
a5 = =
6 6

Therefore, the required terms are
−1 1 1 5 7
, , , , and
6 6 2 6 6

Page : 180 , Block Name : Exercise 9.1

Q5 Write the rst ve terms of the sequences whose n term is a
th n−1 n+1
n = (−1) 5

Answer. Substituting n = 1, 2, 3, 4, 5, we obtain
1−1 1+1 2
a1 = (−1) 5 = 5 = 25

2−1 2+1 3
a2 = (−1) 5 = −5 = −125

3−1 3+1 4
a3 = (−1) 5 = 5 = 625

5−1 4+1 5
a4 = (−1) 5 = −5 = −3125

5 5−1 5+1 6
a = (−1) 5 = 5 = 15625

Therefore, the required terms are 25, –125, 625, –3125, and 15625.

Page : 180 , Block Name : Exercise 9.1

2

Q6 Write the rst ve terms of the sequences whose n term is a
th n +5
n = n
4

Answer. Substituting n = 1, 2, 3, 4, 5, we obtain

2
1 + 5 6 3
a1 = 1 ⋅ = =
4 4 2
2
2 + 5 9 9
a2 = 2 ⋅ = 2 ⋅ =
4 4 2
2
3 + 5 14 21
a3 = 3 ⋅ = 3 ⋅ =
4 4 2
2
4 + 5
a4 = 4 ⋅ = 21
4
2
5 + 5 30 75
a5 =5 ⋅ = 5 ⋅ =
4 4 2

Therefore, the required terms are 3
,
9
,
21
, 21, and
75
.
2 2 2 2

Page : 180 , Block Name : Exercise 9.1

Q7 Find the 17 term in the following sequence whose n term is a
th th
n = 4n − 3; a17 , a24

Answer. Substituting n = 17, we obtain
a17 = 4(17) − 3 = 68 − 3 = 65

Substituting n = 24, we obtain
a24 = 4(24) − 3 = 96 − 3 = 93

Page : 180 , Block Name : Exercise 9.1

Q8 Find the 7 term in the following sequence whose n term is a
2
th th n
n = n
; a7
2

Answer. Substituting n = 7, we obtain
2
7 7
a7 = =
2×7 2

Here, an=n22n.
Substituting n = 7, we obtain
2
7 49
a7 = 7
=
128
2

Page : 180 , Block Name : Exercise 9.1

Q9 Find the 9 th
term in the following sequence whose n th
term is a n = (−1)
n−1 3
n ; a9

Answer. Substituting n = 9, we obtain
9−1 3 3
a9 = (−1) (9) = (9) = 729

Page : 180 , Block Name : Exercise 9.1

n(n−2)
Q10 a n =
n+3
; a20

Answer. Substituting n = 20, we obtain
20(20−2) 20(18) 360
a20 = = =
20+3 23 23

Page : 180 , Block Name : Exercise 9.1

Q11 Write the rst ve terms of the following sequence and obtain the corresponding series:
a1 = 3, an = 3an−1 + 2 for all n > 1

Answer.
a1 = 3, an = 3an−1 + 2 for all n > 1

⇒ a2 = 3a1 + 2 = 3(3) + 2 = 11

a3 = 3a2 + 2 = 3(11) + 2 = 35

a4 = 3a3 + 2 = 3(35) + 2 = 107

a5 = 3a4 + 2 = 3(107) + 2 = 323

Hence, the rst ve terms of the sequence are 3, 11, 35, 107, and 323.
The corresponding series is 3 + 11 + 35 + 107 + 323 + …

Page : 181 , Block Name : Exercise 9.1

Q12 Write the rst ve terms of the following sequence and obtain the corresponding series:
an−1
a1 = −1, an = ,n ≥ 2
n

Answer.
an−1
a1 = −1, an = ,n ≥ 2
n

a1 −1
⇒ a2 = =
2 2

a2 −1
a3 = =
3 6

a3 −1
a4 = =
4 24

a4 −1
a5 = =
4 120

Hence, the rst ve terms of the sequence are
−1 −1 −1 −1
−1, , , , and
2 6 24 120

The corresponding series is
−1 −1 −1 −1
(−1) + ( ) + ( ) + ( ) + ( ) + …
2 6 24 120

Page : 180 , Block Name : Exercise 9.1

Q13 Write the rst ve terms of the following sequence and obtain the corresponding series:
a1 = a2 = 2, an = an−1 − 1, n > 2

Answer.
a1 = a2 = 2, an = an−1 − 1, n > 2

⇒ a3 = a2 − 1 = 2 − 1 = 1

a4 = a3 − 1 = 1 − 1 = 0

a5 = a4 − 1 = 0 − 1 = −1

Hence, the rst ve terms of the sequence are 2, 2, 1, 0, and –1.
The corresponding series is 2 + 2 + 1 + 0 + (–1) + …

Page : 180 , Block Name : Exercise 9.1

Q14 The Fibonacci sequence is de ned by
1 = a1 = a2 and an = an−1 + an−2 , n > 2
an+1
Find , for n = 1, 2, 3, 4, 5
an

Answer.
1 = a1 = a2

an = an−1 + an−2 , n > 2

∴ a3 = a2 + a1 = 1 + 1 = 2

a4 = a3 + a2 = 2 + 1 = 3

a5 = a4 + a3 = 3 + 2 = 5

a6 = a5 + a4 = 5 + 3 = 8
an +1 a2 1
∴ For n = 1, = = = 1
an a1 1

an +1 a3 2
For n = 2, = = = 2
an a1 1

an +1 a4 3
For n = 3, = =
an a3 2

an +1 a5 5
For n = 4, = =
an a4 3

an +1 a6 8
For n = 5, = =
an a5 5

Page : 180 , Block Name : Exercise 9.1

Exercise 9.2

Q1 Find the sum of odd integers from 1 to 2001.

Answer. The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.
This sequence forms an A.P.
Here, rst term, a = 1
Common difference, d = 2
Here, a + (n − 1)d = 2001

⇒ 1 + (n − 1)(2) = 2001

⇒ 2n − 2 = 2000

⇒ n = 1001
n
Sn = [2a + (n − 1)d]
2

1001
∴ Sn = [2 × 1 + (1001 − 1) × 2]
2
1001
= [2 + 1000 × 2]
2
1001
= × 2002
2

= 1001 × 1001

= 1002001

Thus, the sum of odd numbers from 1 to 2001 is 1002001.

Page : 185 , Block Name : Exercise 9.2

Q2 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

Answer. The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, …
995.
Here, a = 105 and d = 5

a + (n − 1)d = 995

⇒ 105 + (n − 1)5 = 995

⇒ (n − 1)5 = 995 − 105 = 890

⇒ n − 1 = 178

⇒ n = 179

179
∴ Sn = [2(105) + (179 − 1)(5)]
2
179
= [2(105) + (178)(5)]
2

= 179[105 + (89)5]

= (179)(105 + 445)

= (179)(550)

= 98450

Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

Page : 185 , Block Name : Exercise 9.2

Q3 In an A.P., the rst term is 2 and the sum of the rst ve terms is one-fourth of the next ve terms.
Show that 20 term is –112.
th

Answer. First term = 2
Let d be the common difference of the A.P.
Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d, …
Sum of rst ve terms = 10 + 10d
Sum of next ve terms = 10 + 35d
According to the given condition,

1
10 + 10d = (10 + 35d)
4

⇒ 40 + 40d = 10 + 35d

⇒ 30 = −5d

⇒ d = −6

∴ a20 = a + (20 − 1)d = 2 + (19)(−6) = 2 − 114 = −112

Thus, the 20th term of the A.P. is –112.

Page : 185 , Block Name : Exercise 9.2

Q4 How many terms of the A.P. −6, − are needed to give the sum –25?
11
, −5, …
2

Answer. Let the sum of n terms of the given A.P. be –25.
It is known that, S = [2a + (n − 1)d],
n
n
2

where n = number of terms, a = rst term, and d = common difference Here, a = –6
11 −11+12 1
d = − + 6 = =
2 2 2

Therefore, we obtain
n 1
−25 = [2 × (−6) + (n − 1) ( )]
2 2

n 1
⇒ −50 = n [−12 + − ]
2 2

25 n
⇒ −50 = n [− + ]
2 2

⇒ −100 = n(−25 + n)

2
⇒ n − 55n + 100 = 0

2
⇒ n − 5n − 20(n − 5) = 0

⇒ n = 20 or 5

Page : 185 , Block Name : Exercise 9.2

Q5 In an A.P., if pth term is and qth term is , prove that the sum of rst pq terms is
1 1

q p

1
(pq + 1), where p ≠ q
2

Answer. It is known that the general term of an A.P. is an = a + (n – 1)d
∴ According to the given information,
th 1
p term = ap = a + (p − 1)d = . . . (1)
q

th 1
q term = aq = a + (q − 1)d = . . . (2)
p

Subtracting (2) from (1), we obtain
1 1
(p − 1)d − (q − 1)d = −
q p

p−q
⇒ (p − 1 − q + 1)d =
pq

p−q
⇒ (p − q)d =
pq

1
⇒ d =
pq

Putting the value of d in (1), we obtain

1 1
a + (p − 1) =
pq q

1 1 1 1
⇒ a = − + =
q q pq pq

pq
∴ Spq = [2a + (pq − 1)d]
2

pq 2 1
= [ + (pq − 1) ]
2 pq pq

1
= 1 + (pq − 1)
2
1 1 1 1
= pq + 1 − = pq +
2 2 2 2
1
= (pq + 1)
2

Thus, the sum of rst pq terms of the A.P. is .
1
(pq + 1)
2

Page : 185 , Block Name : Exercise 9.2

Q6 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

Answer. Let the sum of n terms of the given A.P. be 116.
n
Sn = [2a + (n − 1)d]
2

Here, a = 25 and d = 22 – 25 = – 3
n
∴ Sn = [2 × 25 + (n − 1)(−3)]
2

n
⇒ 116 = [50 − 3n + 3]
2

2
⇒ 232 = n(53 − 3n) = 53n − 3n

2
⇒ 3n − 53n + 232 = 0

2
⇒ 3n − 24n − 29n + 232 = 0

⇒ 3n(n − 8) − 29(n − 8) = 0

⇒ (n − 8)(3n − 29) = 0

29
⇒ n = 8 or n =
3

However, n cannot be equal to . Therefore, n = 8
29

9

∴ a8 = Last term = a + (n – 1)d = 25 + (8 – 1) (– 3)
= 25 + (7) (– 3) = 25 – 21
=4
Thus, the last term of the A.P. is 4.

Page : 185 , Block Name : Exercise 9.2

Q7 Find the sum to n terms of the A.P., whose k th
term is 5k + 1.

Answer. It is given that the k term of the A.P. is 5k + 1.
th

k
th
term = ak = a + (k – 1)d
∴ a + (k – 1)d = 5k + 1
a + kd – d = 5k + 1
Comparing the coef cient of k, we obtain d = 5
a–d=1

⇒a–5=1
⇒a=6
n
Sn = [2a + (n − 1)d]
2
n
= [2(6) + (n − 1)(5)]
2
n
= [12 + 5n − 5]
2
n
= (5n + 7)
2

Page : 185 , Block Name : Exercise 9.2

Q8 If the sum of n terms of an A.P. is (pn + qn ), 2

where p and q are constants, nd the common difference.

Answer. It is known that, S
n
n = [2a + (n − 1)d]
2

According to the given condition,
n 2
[2a + (n − 1)d] = pn + qn
2

n 2
⇒ [2a + nd − d] = pn + qn
2

2 d d 2
⇒ na + n − n ⋅ = pn + qn
2 2

Comparing the coef cients of n2 on both sides, we obtain
d
= q
2

∴d=2q
Thus, the common difference of the A.P. is 2q.

Page : 185 , Block Name : Exercise 9.2

Q9 The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of
their 18 terms. th

Answer. Let a , a , d d be the rst terms and the common difference of the rst and second
1 2 1 2

arithmetic progression respectively.
According to the given condition,
Sum of n terms of first A.P. 5n+4
=
Sum of n terms of second A.P. 9n+6
n
[2a1 +(n−1)d1 ]
2 5n+4
⇒ n
=
[2a2 +(n−1)d2 ] 9n+6
2

2a1 +(n−1)d1 5n+4
⇒ = . . . (1)
2a2 +(n−1)d2 9n+6

Substituting n = 35 in (1), we obtain
2a1 +34d1 5(35)+4
=
2a2 +34d2 9(35)+6

a1 +17d1 179
⇒ = . . . (2)
a2 +17d2 321

th
18 term of first A.P. a1 +17d1
= … (3)
th
18 term of second A.P. a2 +17d2

From (2) and (3), we obtain

th
18 term of first A.P. 179
=
th 321
18 term of second A.P.

Thus, the ratio of 18th term of both the A.P.s is 179: 321.

Page : 185 , Block Name : Exercise 9.2

Q10 If the sum of rst p terms of an A.P. is equal to the sum of the rst q terms, then nd the sum of
the rst (p + q) terms.

Answer. Let a and d be the rst term and the common difference of the A.P. respectively.
Here,
p
Sp = [2a + (p − 1)d]
2
q
Sq = [2a + (q − 1)d]
2

According to the given condition,
p q
[2a + (p − 1)d] = [2a + (q − 1)d]
2 2

⇒ p[2a + (p − 1)d] = q[2a + (q − 1)d]

⇒ 2ap + pd(p − 1) = 2aq + qd(q − 1)

⇒ 2a(p − q) + d[p(p − 1) − q(q − 1)] = 0

2 2
⇒ 2a(p − q) + d [p − p − q + q] = 0

⇒ 2a(p − q) + d[(p − q)(p + q) − (p − q)] = 0

⇒ 2a(p − q) + d[(p − q)(p + q − 1)] = 0

⇒ 2a + d(p + q − 1) = 0

−2a
⇒ d = . . . (1)
p + q − 1

p + q
∴ Sp+q = [2a + (p + q − 1) ⋅ d]
2
p + q −2a
⇒ Sp+q = [2a + (p + q − 1) ( )] [ From (1)]
2 p + q − 1

p + q
= [2a − 2a]
2

= 0

Thus, the sum of the rst (p + q) terms of the A.P. is

Page : 185 , Block Name : Exercise 9.2

Q11 Sum of the rst p, q and r terms of an A.P. are a, b and c, respectively.
Prove that (q − r) + (r − p) + (p − q) = 0
a

p
b

q
c

r

Answer. Let a1 and d be the rst term and the common difference of the A.P. respectively.
According to the given information,

p
Sp = [2a1 + (p − 1)d] = a
2

2a
⇒ 2a1 + (p − 1)d = . . . (1)
p

q
Sq = [2a1 + (q − 1)d] = b
2

2b
⇒ 2a1 + (q − 1)d = . . . (2)
q

r
Sr = [2a1 + (r − 1)d] = c
2

2c
⇒ 2a1 + (r − 1)d = … (3)
r

Subtracting (2) from (1), we obtain
p – 1d – q – 1d = 2ap – 2bq
⇒dp – 1 – q + 1 = 2aq – 2bppq
⇒dp – q = 2aq – 2bppq
⇒d = 2aq – bppqp – q …….(4)
Subtracting (3) from (2), we obtain
2b 2c
(q − 1)d − (r − 1)d = −
q r

2b 2c
⇒ d(q − 1 − r + 1) = −
q r

2br−2qc
⇒ d(q − r) =
qr

2(br−qc)
⇒ d = . . . (5)
qr(q−r)

Equating both the values of d obtained in (4) and (5), we obtain
aq – bppqp – q = br – qcqrq – r
⇒aq – bppp – q = br – qcrq – r
⇒rq – raq – bp = pp – qbr – qc
⇒raq – bpq – r = pbr – qcp – q
⇒aqr – bprq – r = bpr – cpqp – q
Dividing both sides by pqr, we obtain
a b b c
( − ) (q − r) = ( − ) (p − q)
p q q r

a b c
⇒ (q − r) − (q − r + p − q) + (p − q) = 0
p q r

a b c
⇒ (q − r) + (r − p) + (p − q) = 0
p q r

Thus, the given result is proved.

Page : 185 , Block Name : Exercise 9.2

Q12 The ratio of the sums of m and n terms of an A.P. is m : n . 2 2

Show that the ratio of m and n term is (2m – 1) : (2n – 1).th th

Answer. Let a and b be the rst term and the common difference of the A.P. respectively.
According to the given condition,
2
Sum of m terms m
=
Sum of n terms 2
n
m
[2a+(m−1)d] 2
2 m
⇒ n
=
2
⇒ [2a+(n−1)d] n
2

2a+(m−1)d m
⇒ = . . . (1)
2a+(n−1)d n

Putting m = 2m – 1 and n = 2n – 1 in (1), we obtain

2a+(2m−2)d 2m−1
=
2a+(2n−2)d 2n−1

a+(m−1)d 2m−1
⇒ = . . . (2)
a+(n−1)d 2n−1

th a+(m−1)d
m term of A.P.
= . . . (3)
th
n term of A.P. a+(n−1)d

From (2) and (3), we obtain
th
m term of A.P 2m−1
=
th 2n−1
n term of A.P

Thus, the given result is proved.

Page : 185 , Block Name : Exercise 9.2

Q13 If the sum of n terms of an A.P. is 3n 2
+ 5n and its mth term is 164, nd the value of m.

Answer. Let a and b be the rst term and the common difference of the A.P. respectively.
am = a + (m – 1)d = 164 … (1)
Sum of n terms, S = [2a + (n − 1)d] n
n

2

Here,
n 2
[2a + nd − d] = 3n + 5n
2

2 d 2
⇒ na + n ⋅ = 3n + 5n
2

Comparing the coef cient of n2 on both sides, we obtain
d
= 3
2

⇒ d = 6

Comparing the coef cient of n on both sides, we obtain
d
a − = 5
2

⇒ a − 3 = 5

⇒ a = 8

Therefore, from (1), we obtain
8 + (m – 1) 6 = 164
⇒ (m – 1) 6 = 164 – 8 = 156
⇒ m – 1 = 26
⇒ m = 27
Thus, the value of m is 27.

Page : 185 , Block Name : Exercise 9.2

Q14 Insert ve numbers between 8 and 26 such that the resulting sequence is an A.P.

Answer. Let A , A , A , A , andA be ve numbers between 8 and 26 such that
1 2 3 4 5

8, A , A , A , A , A , 26 is an A.P.
1 2 3 4 5

Here, a = 8, b = 26, n = 7
Therefore, 26 = 8 + (7 – 1) d
⇒ 6d = 26 – 8 = 18
⇒d=3
A1 = a + d = 8 + 3 = 11
A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17
A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20
A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23
Thus, the required ve numbers between 8 and 26 are 11, 14, 17, 20, and 23.

Page : 185 , Block Name : Exercise 9.2

n n

Q15 If is the A.M. between a and b, then nd the value of n.
a +b

an−1 +bn−1

Answer. A.M. of a and b =
a+b

2

According to the given condition,
n n
a+b a +b
= n−1 n−1
2 a +b

n−1 n−1 n n
⇒ (a + b) (a + b ) = 2 (a + b )

n n−1 n−1 n n n
⇒ a + ab + ba + b = 2a + 2b

n−1 n−1 n n
⇒ ab + a b = a + b

n−1 n n n−1
⇒ ab − b = a − a b
n−1 n−1
⇒ b (a − b) = a (a − b)

n−1 n−1
⇒ b = a

n−1 0
a a
⇒ ( ) = 1 = ( )
b b

⇒ n − 1 = 0

⇒ n = 1

Page : 185 , Block Name : Exercise 9.2

Q16 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 7 and(m– 1) numbers is 5 : 9. Find the value of m.
th th

Answer. Let A , A , … A be m numbers such that 1, A , A , … A , 31 is an A.P.
1 2 m 1 2 m

Here, a = 1, b = 31, n = m + 2
∴ 31 = 1 + (m + 2 – 1) (d)
⇒ 30 = (m + 1) d
30
⇒ d = . . . (1)
m+1

A1 = a + d
A2 = a + 2d
A3 = a + 3d …
∴ A7 = a + 7d
Am–1 = a + (m – 1) d
According to the given condition,
a+7d 5
=
a+(m−1)d 9

30
1+7( )
(m+1) 5
⇒ = [ From (1)]
30 9
1+(m−1)( )
m+1

m+1+7(30) 5
⇒ =
m+1+30(m−1) 9

m+1+210 5
⇒ =
m+1+30m−30 9

m+211 5
⇒ =
31m−29 9

⇒ 9m + 1899 = 155m − 145

⇒ 155m − 9m = 1899 + 145

⇒ 146m = 2044

⇒ m = 14

Thus, the value of m is 14.

Page : 185 , Block Name : Exercise 9.2

Q17 A man starts repaying a loan as rst instalment of Rs. 100. If he increases the
instalment by Rs 5 every month, what amount he will pay in the 30th instalment?

Answer. The rst installment of the loan is Rs 100.
The second installment of the loan is Rs 105 and so on.
The amount that the man repays every month forms an A.P.
The A.P. is 100, 105, 110, …
First term, a = 100
Common difference, d = 5
A30 = a + (30 – 1)d
= 100 + (29) (5)
= 100 + 145
= 245
Thus, the amount to be paid in the 30th installment is Rs 245.

Page : 186 , Block Name : Exercise 9.2

Q18 The difference between any two consecutive interior angles of a polygon is 5°.
If the smallest angle is 120° , nd the number of the sides of the polygon.

Answer. The angles of the polygon will form an A.P. with common difference d as 5° and rst term a as
120°.
It is known that the sum of all angles of a polygon with n sides is 180° (n – 2).
∘
∴ Sn = 180 (n − 2)
n ∘
⇒ [2a + (n − 1)d] = 180 (n − 2)
2

n ∘ ∘
⇒ [240 + (n − 1)5 ] = 180(n − 2)
2

⇒ n[240 + (n − 1)5] = 360(n − 2)

2
⇒ 240n + 5n − 5n = 360n − 720

2
⇒ 5n + 235n − 360n + 720 = 0

2
⇒ 5n − 125n + 720 = 0

2
⇒ n − 25n + 144 = 0

2
⇒ n − 16n − 9n + 144 = 0

⇒ n(n − 16) − 9(n − 16) = 0

⇒ (n − 9)(n − 16) = 0

⇒ n = 9 or 16

Page : 186 , Block Name : Exercise 9.2

Exercise 9.3

Q1 Find the 20 th
and\(n
th
terms of the G.P. 5

2
,
5

4
,
5

8
,…

Answer. The given G.P. is
5 5 5
, , ,…
2 4 8

Here, a = First term =
5

2
5

r = Common ratio =
4 1
=
5 2
2

19
20−1 5 1 5 5
a20 = ar = ( ) = =
19 20
2 2 (2)(2) (2)

n−1
n−1 5 1 5 5
an = ar = ( ) = =
n−1 n
2 2 (2)(2) (2)

Page : 192 , Block Name : Exercise 9.3

Q2 Find the 12 th
term of a G.P. whose 8 th
term is 192 and the common ratio is 2.

Answer. Common ratio, r = 2
Let a be the rst term of the G.P.
∴ a8 = ar 8–1 = ar7
⇒ ar7 = 192
a(2)7 = 192
a(2)7 = (2)6 (3)
6
(2) ×3 3
⇒ a = =
7
(2) 2

12−1 3 11 10
∴ a12 = ar = ( ) (2) = (3)(2) = 3072
2

Page : 192 , Block Name : Exercise 9.3

Q3 The 5 , 8 and 11
th th th
terms of a G.P. are p, q and s, respectively.
Show that q = ps. 2

Answer. Let a be the rst term and r be the common ratio of the G.P.
According to the given condition,
a5 = a r5–1 = a r4 = p … (1)
a8 = a r8–1 = a r7 = q … (2)
a11 = a r11–1 = a r10 = s … (3)
Dividing equation (2) by (1), we obtain
7 q
ar
=
ar4 p

q
3
r = . . . (4)
p

Dividing equation (3) by (2), we obtain
10
ar s
=
7
ar q
s
3
⇒ r = . . . (5)
q

Equating the values of r3 obtained in (4) and (5), we obtain
q s
=
p q

2
⇒ q = ps

Thus, the given result is proved.

Page : 192 , Block Name : Exercise 9.3

Q4 The 4 th
term of a G.P. is square of its second term, and the rst term is – 3.Determine its 7 th
term.

Answer. Let a be the rst term and r be the common ratio of the G.P.
∴ a = –3
n−1
It is known that, an = ar

3 3
∴ a4 = ar = (−3)r

2
a2 = ar = (−3)r

According to the given condition,

3 2
(−3)r = [(−3)r]

2
⇒ r = 9r

⇒ r = −3

7−1 6 6 7
a7 = ar = ar = (−3)(−3) = −(3) = −2187

Thus, the seventh term of the G.P. is –2187.

Page : 192 , Block Name : Exercise 9.3

Q5 Which term of the following sequences:
(a) 2, 2√2, 4, … is 128?
(b) √3, 3, 3√3, … . is729?
(c) 1

3
,
1

9
,
1

27
, … is
19683
1
?

Answer.
(a) The given sequence is 2, 2√2, 4, …
2√2
Here, a = 2 and r = 2
= √2

Let the nth term of the given sequence be 128.
n−1
an = ar

n−1
⇒ (2)(√2) = 128

n−1
7
⇒ (2)(2) 2 = (2)

n−1
+1 7
⇒ (2) 2 = (2)

n−1
∴ = 6
2

⇒ n − 1 = 12

⇒ n = 13

Thus, the 13th term of the given sequence is 128.
(b) The given sequence is √3, 3, 3√3
Here, a = √3 and r = = √3
3

√3

Let the nth term of the given sequence be 729.
n−1
an = ar

n−1
∴ ar = 729

n−1
⇒ (√3)(√3) = 729

1 n−1
6
⇒ (3) 2 (3) 2 = (3)

1 n−1
+ 6
⇒ (3) 2 2 = (3)
1 n−1
∴ + = 6
2 2

1+n−1
⇒ = 6
2

⇒ n = 12

Thus, the 12th term of the given sequence is 729.
(c) The given sequence is , , , … 1

3
1

9 27
1

Here, a = 1

3
and r =
1

9
÷
1

3
=
1

3

Let the nth term of the given sequence be 1

19683
.
n−1
an = ar

n−1 1
∴ ar =
19683

n−1
1 1 1
⇒ ( )( ) =
3 3 19683

n 9
1 1
⇒ ( ) = ( )
3 3

⇒ n = 9

Thus, the 9th term of the given sequence is 19683
1
.

Page : 192 , Block Name : Exercise 9.3

Q6 For what values of x, the numbers − are in G.P.?
2 7
, x, − a
7 2

Answer. The given numbers are − .
2 7
, x, − a
7 2
x

Common ratio =
−2 −7x
=
−2 2
7

−7

Also, common ratio = 2 −7
=
x 2x
−7x −7
∴ =
2 2x

2 −2×7
⇒ x = = 1
−2×7

⇒ x = √1

⇒ x = ±1

Thus, for x = ± 1, the given numbers will be in G.P.

Page : 192 , Block Name : Exercise 9.3

Q7 Find the sum to n terms in the geometric progression 0.15, 0.015, 0.0015, ... 20 terms.

Answer. The given G.P. is 0.15, 0.015, 0.00015, …
Here, a = 0.15 and r =
0.015
= 0.1
0.15
n
a (1 − r )
Sn =
1 − r
20
0.15 [1 − (0.1) ]
∴ S20 =
1 − 0.1

0.15
20
= [1 − (0.1) ]
0.9

15
20
= [1 − (0.1) ]
90

1
20
= [1 − (0.1) ]
6

Page : 192 , Block Name : Exercise 9.3

Q8 Find the sum to n terms in the geometric progression √7, √21, 3√7, … , n

Answer.
The given G.P. is √7, √21, 3√7, …

Here, a = √7

√21
r = = √3
√7

n
a(1−r )
Sn =
1−r

n
√7 [1 − (√3) ]

∴ Sn =
1 − √3

n
√7 [1 − (√3) ]
1 + √3
= × [by rationalizing]
1 − √3 1 + √3

n
√7(1 + √3) [1 − (√3) ]

=
1 − 3

−√7(1 + √3) n

= [1 − (3) 2 ]
2

√7(1 + √3) n

= [(3) 2 − 1]
2

Page : 192 , Block Name : Exercise 9.3

Q9
Find the sum to n terms in the geometric progression

2 3
1, −a, a , −a … ( if a ≠ −1)

Answer.
2 3
The given G.P. is 1, −a, a , −a , … … … …

Here, first term = a1 = 1

Common ratio = r = −a
n
a1 (1−r )
Sn =
1−r

n n
1[1−(−a) ] [1−(−a) ]
∴ Sn = =
1−(−a) 1+a

Page : 192 , Block Name : Exercise 9.3

Q10 Find the sum to n terms in the geometric progression x , x , x , … n terms (if x ≠ ±1) 3 5 7

Answer.
3 5 7
The given G.P. is x , x , x , …

3 2
Here, a = x and r = x

n
3 2
n x [1−(x ) ] 3 2n
a(1−r ) x (1−x )
Sn = = =
1−r 1−x2 1−x2

Page : 192 , Block Name : Exercise 9.3

Q11 Evaluate ∑
11 k
(2 + 3 )
k=1

Answer.
11 k 11 11 k 11 k 11 k
∑ (2 + 3 ) = ∑ (2) + ∑ 3 = 2(11) + ∑ 3 = 22 + ∑ 3 … (1)
k=1 k=1 k=1 k=1 k=1

11 k 1 2 3 11
∑ 3 = 3 + 3 + 3 + … + 3
k=1
2 3
The terms of this sequence 3, 3 , 3 , … forms a G.P.
n
a(r −1)
Sn =
r−1

11
3[(3) −1]

⇒ Sn =
3−1

3 11
⇒ Sn = (3 − 1)
2

11 k 3 11
∴ ∑ 3 = (3 − 1)
k=1 2

Substituting this value in equation (1), we obtain

11 k 3 11
∑ (2 + 3 ) = 22 + (3 − 1)
k=1 2

Page : 192 , Block Name : Exercise 9.3

Q12 The sum of rst three terms of a G.P. is and their product is 1.
39

10

Find the common ratio and the terms.

Answer.
a
,a,ar
Let r r be the first three terms of the G.P.

a 39
+ a + ar = . . . (1)
r 10

a
( ) (a)(ar) = 1 . . . (2)
r

From (2), we obtain
3
a = 1

⇒ a = 1 (Considering real roots only)
Substituting a = 1 in equation (1), we obtain
1 39
+ 1 + r =
r 10

2 39
⇒ 1 + r + r = r
10

2
⇒ 10 + 10r + 10r − 39r = 0

2
⇒ 10r − 29r + 10 = 0

2
⇒ 10r − 25r − 4r + 10 = 0

⇒ 5r(2r − 5) − 2(2r − 5) = 0

⇒ (5r − 2)(2r − 5) = 0

2 5
⇒ r = or
5 2

Thus, the three terms of G.P. are 5

2
, 1, and
2

5

Page : 192 , Block Name : Exercise 9.3

Q13 How many terms of G.P.3, 3 , 3 , … are needed to give the sum 120? 2 3

Answer. The given G.P. is 3, 3 , 3 , …
2 3

Let n terms of this G.P. be required to obtain the sum as 120.
n
a(r −1)
Sn =
r−1

Here, a = 3 and r = 3

n
3(3 −1)
∴ Sn = 120 =
3−1
n
3(3 −1)
⇒ 120 =
2

120×2 n
⇒ = 3 − 1
3

n
⇒ 3 − 1 = 80
n
⇒ 3 = 81

∴ n = 4

Thus, four terms of the given G.P. are required to obtain the sum as 120.

Page : 192 , Block Name : Exercise 9.3

Q14 The sum of rst three terms of a G.P. is 16 and the sum of the next three terms is 128.
Determine the rst term, the common ratio and the sum to n terms of the G.P.

Answer.
2 3
Let the G.P. be a, ar, ar , ar , …

According to the given condition,

2 3 4 5
a + ar + ar = 16 and ar + ar + ar = 128
2
⇒ a (1 + r + r ) = 16 … (1)

3 2
ar (1 + r + r ) = 128 … (2)

Dividing equation (2) by (1), we obtain
3 2
ar (1+r+r )
128
2
=
a(1+r+r ) 16

3
⇒ r = 8

∴ r = 2

Substituting r = 2 in (1), we obtain
a (1 + 2 + 4) = 16<
⇒ a (7) = 16
16
⇒ a =
7

n
a(r −1)
Sn =
r−1

n
16 (2 −1) 16 n
⇒ Sn = = (2 − 1)
7 2−1 7

Page : 192 , Block Name : Exercise 9.3

Q15 Given a G.P. with a = 729 and 7th term 64, determine S7 .

Answer. a = 729
a7 = 64
Let r be the common ratio of the G.P.
It is known that,a = ar n
n−1

7−1 6
a7 = ar = (729)r

6
⇒ 64 = 729r

6 64
⇒ r =
729

6
6 2
⇒ r = ( )
3

2
⇒ r =
3
n
a(1−J )
Also, it is known that, S n =
1−r

7
2
729 [1 − ( ) ]
3

∴ S7 =
2
1 −
3

7
2
= 3 × 729 [1 − ( ) ]
3

7 7
(3) − (2)
7
= (3) − ( ]
7
(3)

= 2187 − 128

= 2059

Page : 192 , Block Name : Exercise 9.3

Q16 Find a G.P. for which sum of the rst two terms is – 4 and the fth term is 4 times the third term.

Answer. Let a be the rst term and r be the common ratio of the G.P.
According to the given conditions,
2
a(1−r )
S2 = −4 = . . . (1)
1−r

a5 = 4 × a3

4 2
a = 4 ar
5

2
⇒ r = 4

∴ r = ±2

From (1), we obtain
2
a[1−(2) ]
−4 = for r = 2
1−2

a(1−4)
⇒ −4 =
−1

⇒ −4 = a(3)

−4
⇒ a =
3
2
a[1−(−2) ]
Also, − 4 = for r = −2
1−(−2)

a(1−4)
⇒ −4 =
1+2

a(−3)
⇒ −4 =
3

⇒ a = 4

Thus, the required G.P. is

4,-8,16,-32,...
−4 −8 −16
, , ,…
3 3 3

Page : 192 , Block Name : Exercise 9.3

Q17 If the 4 , 10 and16 terms of a G.P. are x, y and z, respectively.
th th th

Prove that x, y, z are in G.P.

Answer. Let a be the rst term and r be the common ratio of the G.P.
According to the given condition,
3
a4 = ar = x … (1)

9
a10 = ar = y … (2)

15
a16 = ar = z … (3)

Dividing (2) by (1), we obtain
y 9 y
ar 6
= ⇒ = r
x ar
3 x

Dividing (3) by (2), we obtain
15
z ar z 6
= ⇒ = r
y ar9 y

y z
∴ =
x y

Thus, x, y, z are in G. P.

Page : 192 , Block Name : Exercise 9.3

Q18 Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

Answer. The given sequence is 8, 88, 888, 8888…
This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as
Sn = 8 + 88 + 888 + 8888 + …………….. to n terms
8
= [9 + 99 + 999 + 9999 + … … … … to n terms ]
9

8
2 3 4
= [(10 − 1) + (10 − 1) + (10 − 1) + (10 − 1) + … … . to n terms ]
9
8
2
= [(10 + 10 + … . . n terms ) − (1 + 1 + 1 + … n terms )]
9
n
8 10 (10 − 1)
= [ − n]
9 10 − 1
n
8 10 (10 − 1)
= [ − n]
9 9

80 8
n
= (10 − 1) − n
81 9

Page : 193 , Block Name : Exercise 9.3

Q19 Find the sum of the products of the corresponding terms of the sequences 2,4,8,
.
1
16, 32 and 128, 32, 8, 2,
2

Answer.

1
Required sum = 2 × 128 + 4 × 32 + 8 × 8 + 16 × 2 + 32 ×
2

1 1
= 64 [4 + 2 + 1 + + ]
2 2
2

1 1
Here, 4, 2, 1, , is a G.P.
2 2
2

First term, a = 4
1
Common ratio, r =
2
n
a(1−r )
Sn =
1−r

5
1
4[1−( ) ] 1
2 4[1− ]
32 32−1 31
∴ S5 = = = 8( ) =
1 1 32 4
1−
2 2

31
∴ Required sum = 64 ( ) = (16)(31) = 496
4

Page : 193 , Block Name : Exercise 9.3

Q20 Show that the products of the corresponding terms of the sequences
a, ar, ar … ar 2 n−1
and A, AR, AR , … AR form a G.P, and nd the common ratio.
2 n−1

Answer. It has to be proved that the sequence,a, ar, ar … ar 2 n−1 2
and A, AR, AR , … AR
n−1
form a
G.P,
Second term arAR
= = rR
First term aA
2 2
Third term ar AR
= = rR
Second term arAR

Thus, the above sequence forms a G.P. and the common ratio is rR.

Page : 193 , Block Name : Exercise 9.3

Q21 Find four numbers forming a geometric progression in which the third term is greater than the
rst term by 9,
and the second term is greater than the 4th by 18.

Answer. Let a be the rst term and r be the common ratio of the G.P.
2 3
a1 = a, a2 = ar, a3 = ar , a4 = ar

By the given condition,

a3 = a1 + 9

2
⇒ ar = a + 9 … (1)

a2 = a4 + 18

3
⇒ ar = ar + 18 … (2)

From (1) and (2), we obtain

2
a (r − 1) = 9 … (3)

2
ar (1 − r ) = 18 … (4)

Dividing (4) by (3), we obtain

2
ar(1−r )
18
2
=
a(r −1) 9

⇒ −r = 2

⇒ r = −2

Substituting the value of r in (1), we obtain
4a = a + 9
⇒ 3a = 9
∴a=3
Thus, the rst four numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3(–2)3 i.e., 3¸–6, 12, and –24.

Page : 193 , Block Name : Exercise 9.3

Q22 If the p , q andr th th th
terms of a G.P. are a, b and c, respectively.
Prove that a b c q−r r−p P −q
= 1

Answer. Let A be the rst term and R be the common ratio of the G.P.
According to the given information,
p−1
AR = a

q−1
AR = b

r−1
AR = c

q−r r−p p−q
a b c
q−r (p−1)(q−r) r−p (q−1)(r−p) p−q (r−1)(p−q)
= A × R × A × R × A × R
−r+r−p+p−q
= Aq × R(pr − pr − q + r) + (rq − r + p − pq) + (pr − p − qr + q)
0 0
= A × R

= 1

Thus, the given result is proved.

Page : 193 , Block Name : Exercise 9.3

Q23 If the rst and the n th
term of a G.P. are a and b, respectively, and if P is the product of n terms,
prove that P = (ab) . 2 n

Answer. The rst term of the G.P is a and the last term is b.
Therefore, the G.P. is a,ar , ar , … ar where r is the common ratio.
2 3 n−1

n−1
b = ar … (1)

P = Product of n terms
2 n−1
= (a)(ar) (ar ) … (ar )

2 n−1
= (a × a × … a) (r × r × …r )

n 1+2+…(n−1)
= a r … (2)

Here, 1, 2, …(n – 1) is an A.P.
∴1 + 2 + ……….+ (n – 1)
n−1 n−1 n(n−1)
= [2 + (n − 1 − 1) × 1] = [2 + n − 2] =
2 2 2

n(n−1)

n
P = a r 2

2 2n n(n−1)
∴ P = a r

n
2 (n−1)
= [a r ]

n
n−1
= [a × ar ]

n
= (ab) [using(1)]

Thus, the given result is proved.

Page : 193 , Block Name : Exercise 9.3

Q24 Show that the ratio of the sum of rst n terms of a G.P. to the sum of terms from
th th 1
(n + 1) to (2n) term is n
r

Answer. Let a be the rst term and r be the common ratio of the G.P.
n
a(1−r )
Sum of 1st n terms = (1−r)
n
an+1 (1−r )
Since there are n terms from (n +1)th to (2n)th term,= (1−r)
n
a(1−r ) (1−r)
Thus, required ratio = ×
n n
=
1

r
n
(1−r) ar (1−r )

Thus, the ratio of the sum of rst n terms of a G.P. to the sum of terms from (n + 1) th
to (2n) th
term is
1

r
n
.

Page : 193 , Block Name : Exercise 9.3

Q25 If a, b, c and d are in G.P. show that
2 2 2 2 2 2 2
(a + b + c ) (b + c + d ) = (ab + bc + cd)

Answer. a, b, c, d are in G.P. Therefore,
bc = ad … . (1)

2
b = ac … (2)

2
c = bd … (3)

It has to be proved that,

2 2 2 2 2 2 2
(a + b + c ) (b + c + d ) = (ab + bc − cd)

R.H.S.
2
= (ab + bc + cd)

2
= (ab + ad + cd) [U sing(1)]

2
= [ab + d(a + c)]

2 2 2 2
= a b + 2abd(a + c) + d (a + c)

2 2 2 2 2 2 2
= a b + 2a c + 2acbd + d (a + 2ac + c )

2 2 2 2 2 2 2 2 2 2 2 2
= a b + 2a c + 2b c + d a + 2a b + d c [U sing(1) and (2)]
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
= a b + a c + a c + b c + b c + d a + d b + d b
2 2
+ d c
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
= a b + a c + a d + b × b + b c + b d + c b + c × c + c d

[Using (2) and (3) and rearranging terms]

2 2 2 2 2 2 2 2 2 2 2 2
= a (b + c + d ) + b (b + c + d ) + c (b + c + d )

2 2 2 2 2 2
= (a + b + c ) (b + c + d )

= L.H.S.
∴ L.H.S. = R.H.S.
2 2 2 2 2 2 2
∴ (a + b + c ) (b + c + d ) = (ab + bc + cd)

Page : 193 , Block Name : Exercise 9.3

Q26 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Answer.
Let G1 and G2 be two numbers between 3 and 81 such

that the series, 3, G1 , G2 , 81, forms a G.P.

Let a be the rst term and r be the common ratio of the G.P.
3
∴ 81 = (3)(r)

3
⇒ r = 27

∴ r = 3 (Taking real roots only)

For r = 3

G1 = ar = (3)(3) = 9

2 2
G2 = ar = (3)(3) = 27

Thus, the required two numbers are 9 and 27.

Page : 193 , Block Name : Exercise 9.3

n+1 n+1

Q27 Find the value of n so that may be the geometric mean between a and b.
a +b

a +bn
n

Answer.
G. M. of a and b is √ab
n+1 n+1
a +b
By the given condition, n n
= √ab
a +b

Squaring both sides, we obtain
2
n+1 n+1
(a +b )
= ab
n n 2
(a +b )

2n+2 n+1 n+1 2n+2 2n n n 2n
⇒ a + 2a b + b = (ab) (a + 2a b + b )

2n+2 n+1 n+1 2n+1 n+1 n+1 2n+1
⇒ a + 2a b + b b + 2a b + ab

2n+2 2n+1 2n+1
⇒ a + b b + ab
2n+2 2n+1 2n+1 2n+2
⇒ a − a b = ab − b
2n+1 2n+1
⇒ a (a − b) = b (a − b)

2n+1 0
a a
⇒ ( ) = 1 = ( )
b b

⇒ 2n + 1 = 0

−1
⇒ n =
2

Page : 193 , Block Name : Exercise 9.3

Q28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (
(3 + 2√2) : (3 − 2√2)

Answer. Let the two numbers be a and b.
G. M. = √ab

According to the given condition,
a + b = 6√ab

2
⇒ (a + b) = 36(ab) . . . (1)

Also,
2 2
(a − b) = (a + b) − 4ab = 36ab − 4ab = 32ab

⇒ a − b = √32√ab

= 4√2√ab . . . (2)

Adding (1) and (2), we obtain
2a = (6 + 4√2)√ab

⇒ a = (3 + 2√2)√ab

Substituting the value of a in (1), we obtain
b = 6√ab − (3 + 2√2)√ab

⇒ b = (3 − 2√2)√ab

a (3+2√2)√ab 3+2√2
= =
b (3−2√2)√ab 3−2√2

Thus, the required ratio is (3 + 2√2) : (3 − 2√2)

Page : 193 , Block Name : Exercise 9.3

Q29 If A and G be A.M. and G.M., respectively between two positive numbers,
prove that the numbers are A ± √(A + G)(A − G).

Answer. It is given that A and G are A.M. and G.M. between two positive numbers.
Let these two positive numbers be a and b.
a+b
∴ AM = A = … (1)
2

GM = G = √ab … (2)

From (1) and (2), we obtain

a + b = 2A … (3)

2
ab = G … (4)

Substituting the value of a and b from (3) and (4) in the

2 2
identity (a − b) = (a + b) − 4ab, we obtain
2 2 2 2 2
(a − b) = 4A − 4G = 4 (A − G )

2
(a − b) = 4(A + G)(A − G)

(a − b) = 2√(A + G)(A − G) . . . (5)

From (3) and (5), we obtain

2a = 2A + 2√(A + G)(A − G)

⇒ a = A + √(A + G)(A − G)

Substituting the value of a in (3), we obtain

b = 2A − A − √(A + G)(A − G) = A − √(A + G)(A − G)

Thus, the two numbers are A ± √(A + G)(A − G)

Page : 193 , Block Name : Exercise 9.3

Q30 The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in
the culture originally,
how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ?

Answer. It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria
after every hour will form a G.P.
Here, a = 30 and r = 2
Here, a = 30 and r = 2

2 2
∴ a3 = ar = (30)(2) = 120

Therefore, the number of bacteria at the end of 2nd hour will be 120.
th
The number of bacteria at the end of 4 hour will be 480.
n n
an+1 = ar = (30)2

th
Thus, number of bacteria at the end of n hour will be

n
30(2) .

Page : 193 , Block Name : Exercise 9.3

Q31 What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate
of 10% compounded annually?

Answer. The amount deposited in the bank is Rs 500.
At the end of rst year, amount =Rs 500 (1 + 1

10
) = Rs 500 (1.1)
At the end of 2nd year, amount = Rs 500 (1.1) (1.1)
At the end of 3rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on
∴Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)
= Rs 500(1.1) 10

Page : 193 , Block Name : Exercise 9.3

Q32 If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the
quadratic equation.

Answer. Let the root of the quadratic equation be a and b.
According to the given condition,

a + b
A ⋅ M. = = 8 ⇒ a + b = 16 . . . (1)
2

G. M = √ab = 5 ⇒ ab = 25 . . . (2)

The quadratic equation is given by,
2
x − x( sum of roots) + ( Product of roots ) = 0

2
x − x(a + b) + ( Product of roots ) = 0

2
x − 16x + 25 = 0 [using (1) and (2)]

Thus, the required quadratic equation is x 2
− 16x + 25 = 0.

Page : 193 , Block Name : Exercise 9.3

Exercise 9.4

Q1 Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +...

Answer. The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
nth term, an = n ( n + 1)
n n

∴ Sn = ∑ ak = ∑ k(k + 1)

k=1 k=1

n n

2
= ∑ k + ∑ k

k=1 k=1

n(n + 1)(2n + 1) n(n + 1)
= +
6 2

n(n + 1) 2n + 4
= ( )
2 3

2n+4
n(n + 1) ( )
3

=
3

n(n + 1)(n + 2)
=
3

Page : 196 , Block Name : Exercise 9.4

Q2 Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

Answer. The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
nth term, an = n ( n + 1) ( n + 2)

2
= (n + n) (n + 2)

3 2
= n + 3n + 2n
n

∴ Sn = ∑ ak

k=1

n n n

3 2
= ∑ k + 3∑ k + 2∑ k

k=1 k=1 k=1

2
n(n + 1) 3n(n + 1)(2n + 1) 2n(n + 1)
= [ ] + +
2 6 2

2
n(n + 1) n(n + 1)(2n + 1)
= [ ] + + n(n + 1)
2 2

n(n + 1) n(n + 1)
= [ + 2n + 1 + 2]
2 2

2
n(n + 1) n + n + 4n + 6
= [ ]
2 2

n(n + 1)
2
= (n + 5n + 6)
4
n(n + 1)
2
= (n + 2n + 3n + 6)
4

n(n + 1)[n(n + 2) + 3(n + 2)]
=
4

n(n + 1)(n + 2)(n + 3)
=
4

Page : 196 , Block Name : Exercise 9.4

Q3 Find the sum to n terms of the series 3 × 1 2
+ 5 × 2
2
+ 7 × 3
2
+ …

Answer.
2 2 2
The given series is 3 × 1 + 5 × 2 + 7 × 3 + …

th 2 3 2
n term, an = (2n + 1)n = 2n + n

n
∴ Sn = ∑ ak
k=1

n 3 2 n 3 n 2
= ∑ = (2k + k ) = 2∑ k + ∑ k
k=1 k=1 k=1

2
n(n + 1) n(n + 1)(2n + 1)
= 2[ ] +
2 6

2 2
n (n + 1) n(n + 1)(2n + 1)
= +
2 6

n(n + 1) 2n + 1
= [n(n + 1) + ]
2 3

2
n(n + 1) 3n + 3n + 2n + 1
= [ ]
2 3

2
n(n + 1) 3n + 5n + 1
= [ ]
2 3

2
n(n + 1) (3n + 5n + 1)
=
6

Page : 196 , Block Name : Exercise 9.4

Q4 Find the sum to n terms of the series 1×2
1
+
1

2×3
+
1

3×4
+ …

Answer, The given series is
1 1 1
+ + + …
1×2 2×3 3×4

th 1
n term, an =
n(n+1)

1 1
= − [by partial fraction]
n n+1

1 1
a1 = −
1 2

1 1
a2 = −
2 3

1 1
a3 = −
3 3

1 1
a3 = − …
3 4

1 1
an = −
3 4+1

Adding the above terms column wise, we obtain
1 1 1 1 1 1 1 1
a1 + a2 + … + an = [ + + + … ] − [ + + + … ]
1 2 3 n 2 3 4 n+1

1 n+1−1 n
∴ Sn = 1 − = =
n+1 n+1 n+1

Page : 196 , Block Name : Exercise 9.4

Q5 Find the sum to n terms of the series 5 2
+ 6
2
+ 7
2
+ … + 20
2

Answer.
2 2 2 2
The given series is 5 + 6 + 7 + … + 20

th 2 2
n term, an = (n + 4) = n + 8n + 16
n n

2
∴ Sn = ∑ ak = ∑ (k + 8k + 16)

k=1 k=1

n n n

2
= ∑ k + 8 ∑ k + ∑ 16

k=1 k=1 k=1

n(n + 1)(2n + 1) 8n(n + 1)
= + + 16n
6 2

th 2 2
16 term is (16 + 4) = 20 2

16(16 + 1)(2 × 16 + 1) 8 × 16 × (16 + 1)
∴ S th = + + 16 × 16
16
6 2

(16)(17)(33) (8) × 16 × (16 + 1)
= + + 16 × 16
6 2

(16)(17)(33) (8)(16)(17)
= + + 256
6 2
= 1496 + 1088 + 256

= 2840

2 2 2 2
∴ 5 + 6 + 7 + … … . +20 = 2840

Page : 196 , Block Name : Exercise 9.4

Q6 Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 + …

Answer. The given series is 3 × 8 + 6 × 11 + 9 × 14 + …
th th
an = (n term of 3, 6, 9 …) × (n term of 8, 11, 14, …)

= (3n)(3n + 5)

2
= 9n + 15n
n n 2
∴ Sn = ∑ ak = ∑ (9k + 15k)
k=1 k=1

n n

2
= 9∑ k + 15 ∑ k

k=1 k=1

n(n + 1)(2n + 1) n(n + 1)
= 9 × + 15 ×
6 2

3n(n + 1)(2n + 1) 15n(n + 1)
= +
6 2

3n(n + 1)
= (2n + 1 + 5)
2

3n(n + 1)
= (2n + 6)
2

= 3n(n + 1)(n + 3)

Page : 196 , Block Name : Exercise 9.4

Q7 Find the sum to n terms of the series 1 2
+ (1
2 2
+ 2 ) + (1
2
+ 2
2 2
+ 3 ) + …

Answer.

2 2 2 2 2 3
The given series |s1 + (1 + 2 ) + (1 + 2 + 3 ) + …

2 2 3 2
an = (1 + 2 + 3 + …… + n )

n(n+1)(2n+1)
=
6

2
n(2n +3n+1) 3 2
2 +3n +n
= =
6 6

1 3 1 2 1
= n + n + n
3 2 6

n
∴ Sn = ∑ ak
k=1
n
1 1 1
3 2
= ∑ ( k + k + k)
3 2 6
k=1

n n n
1 1 1
3 2
= ∑ k + ∑ k + ∑ k
3 2 6
k=1 k=1 k=1

2 2
1 n (n + 1) 1 n(n + 1)(2n + 1) 1 n(n + 1)
= + × + ×
3 2 6 6
(2) 2 2

n(n + 1) n(n + 1) (2n + 1) 1
= [ + + ]
6 2 2 2

2
n(n + 1) n + n + 2n + 1 + 1
= [ ]
6 2

2
n(n + 1) n + n + 2n + 2
= [ ]
6 2

n(n + 1) n(n + 1) + 2(n + 1)
= [ ]
6 2

n(n + 1) (n + 1)(n + 2)
= [ ]
6 2

2
n(n + 1) (n + 1)(n + 2)
= ]
2
2
n(n + 1) (n + 1)(n + 2)
=
12

Page : 196 , Block Name : Exercise 9.4

Q8 Find the sum to n terms of the series n(n + 1)(n + 4)

Answer.
2 3 2
an =n(n + 1)(n + 4) = n (n + 5n + 4) = n + 5n + 4n

n n n n

3 2
∴ Sn = ∑ ak = ∑ k + 5∑ k + 4∑ k

k=1 k=1 k=1 i=1

2 2
n (n + 1) 5n(n + 1)(2n + 1) 4n(n + 1)
= + +
4 6 2

n(n + 1) n(n + 1) 5(2n + 1)
= [ + + 4]
2 2 3

2
n(n + 1) 3n + 3n + 20n + 10 + 24
= [ ]
2 6

2
n(n + 1) 3n + 23n + 34
= [ ]
2 6

2
n(n + 1) (3n + 23n + 34)
=
12

Page : 196 , Block Name : Exercise 9.4

Q9 Find the sum to n terms of the series n 2 n
+ 2

Answer.
2 n
an = n + 2

n 2 k n 2 n k
∴ Sn = ∑ k + 2 = ∑ k + ∑ 2 . . . (1)
k=1 k−1 k=1

n k 1 2 3
Consider ∑ 2 = 2 + 2 + 2 + …
k=1
2 3
The above series 2, 2 , 2 , … is a G.P. with both the first

term and common ratio equal to 2 .
n
n k (2)[(2) −1] n
∴ ∑ 2 = = 2 (2 − 1) . . . (2)
k=1 2−1

Therefore, from (1) and (2), we obtain

n n n(n+1)(2n+1) n
2
Sn = ∑ k + 2 (2 − 1) = + 2 (2 − 1)
k=1 6

Page : 196 , Block Name : Exercise 9.4

Q10Find the sum to n terms of the series (2n − 1) 2

Answer.
2 2
an = (2n − 1) = 4n − 4n + 1

n n 2
∴ Sn = ∑ ak = ∑ (4k − 4k + 1)
k=1 k=1

n 2 n n
= 4∑ k − 4∑ k + ∑ 1
k=1 k=1 k=1

4n(n+1)(2n+1) 4n(n+1)
= − + n
6 2

2n(n + 1)(2n + 1)
= − 2n(n + 1) + n
3
2
2 (2n + 3n + 1)
= n[ − 2(n + 1) + 1]
3

2
4n + 6n + 2 − 6n − 6 + 3
= n[ ]
3

2
4n − 1
= n[ ]
3

n(2n + 1)(2n − 1)
=
3

n(2n + 1)(2n − 1)
=
3

Page : 196 , Block Name : Exercise 9.4

Miscellaneous Exercise

Q1 Show that the sum of (m + n) th
and (m– n) th
terms of an A.P. is equal to twice the m th
term.

Answer. Let a and d be the rst term and the common difference of the A.P. respectively.
It is known that the kth term of an A. P. is given by
ak = a + (k − 1)d

∴ am−n = a + (m + n − 1)d

am−n = a + (m − n − 1)d

am = a + (m − 1)d

∴ am+n + am−n = a + (m + n − 1)d + a + (m − n − 1)d

= 2a + (m + n − 1 + m − n − 1)d

= 2a + (2m − 2)d

= 2a + 2(m − 1)d

= 2[a + (m − 1)d]

= 2am

Thus, the sum of (m + n) th
and (m– n) th
terms of an A.P. is equal to twice the m th
term.

Page : 199 , Block Name : Miscellaneous Exercise

Q2 If the sum of three numbers in A.P., is 24 and their product is 440, nd the numbers.

Answer. Let the three numbers in A.P. be a – d, a, and a + d.
According to the given information,
(a – d) + (a) + (a + d) = 24 … (1)
⇒ 3a = 24

∴a=8
(a – d) a (a + d) = 440 … (2)
⇒ (8 – d) (8) (8 + d) = 440
⇒ (8 – d) (8 + d) = 55
2
⇒ 64 − d = 55

2
⇒ d = 64 − 55 = 9

⇒ d = ±3

Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5.
Thus, the three numbers are 5, 8, and 11.

Page : 199 , Block Name : Miscellaneous Exercise

Q3 Let the sum of n, 2 , 3 terms of an A.P. be S , S , S , respectively,
n n 1 2 3

show that S = 3 (S − S )
3 2 1

Answer. Let a and b be the rst term and the common difference of the A.P. respectively.
Therefore,
n
S1 = [2a + (n − 1)d] … (1)
2

2n
S2 = [2a + (2n − 1)d] = n[2a + (2n − 1)d] … (2)
2

3n
S3 = [2a + (3n − 1)d] … (3)
2

From (1) and (2), we obtain
n
S2 − S1 = n[2a + (2n − 1)d] − [2a + (n − 1)d]
2

4a + 4nd − 2d − 2a − nd + d
= n{ }
2

2a + 3nd − d
= n[ ]
2

n
= [2a + (3n − 1)d]
2
3n
∴ 3 (S2 − S1 ) = [2a + (3n − 1)d] = S3 [ From (3)]
2

Hence, the given result is proved.

Page : 199 , Block Name : Miscellaneous Exercise

Q4 Find the sum of all numbers between 200 and 400 which are divisible by 7.

Answer. The numbers lying between 200 and 400, which are divisible by 7, are
203, 210, 217, … 399
∴First term, a = 203
Last term, l = 399
Common difference, d = 7
Let the number of terms of the A.P. be n.
∴ an = 399 = a + (n –1) d
⇒ 399 = 203 + (n –1) 7
⇒ 7 (n –1) = 196
⇒ n –1 = 28

⇒ n = 29
29
∴ S29 = (203 + 399)
2
29
= (602)
2

= (29)(301)

= 8729

Thus, the required sum is 8729

Page : 199 , Block Name : Miscellaneous Exercise

Q5 Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Answer. The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.
This forms an A.P. with both the rst term and common difference equal to 2.
⇒100 = 2 + (n –1) 2
⇒ n = 50
50
∴ 2 + 4 + 6 + … + 100 = [2(2) + (50 − 1)(2)]
2
50
= [4 + 98]
2

= (25)(102)

= 2550

The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.
This forms an A.P. with both the rst term and common difference equal to 5.
∴100 = 5 + (n –1) 5
⇒ 5n = 100
⇒ n = 20
20
∴ 5 + 10 + … + 100 = [2(5) + (20 − 1)5]
2

= 10[10 + (19)5]

= 10[10 + 95] = 10 × 105

= 1050

The integers, which are divisible by both 2 and 5, are 10, 20, … 100.
This also forms an A.P. with both the rst term and common difference equal to 10.
∴100 = 10 + (n –1) (10)
⇒ 100 = 10n
⇒ n = 10
10
∴ 10 + 20 + … + 100 = [2(10) + (10 − 1)(10)]
2

= 5[20 + 90] = 5(110) = 550

∴Required sum = 2550 + 1050 – 550 = 3050
Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

Page : 199 , Block Name : Miscellaneous Exercise

Q6 Find the sum of all two digit numbers which when divided by 4, yields 1 as Remainder.

Answer. The two-digit numbers, which when divided by 4, yield 1 as remainder, are
13, 17, … 97.
This series forms an A.P. with rst term 13 and common difference 4.
Let n be the number of terms of the A.P.
It is known that the nth term of an A.P. is given by, an = a + (n –1) d
∴97 = 13 + (n –1) (4)
⇒ 4 (n –1) = 84
⇒ n – 1 = 21
⇒ n = 22
Sum of n terms of an A.P. is given by,
n
Sn = [2a + (n − 1)d]
2
22
∴ S22 = [22(13) + (22 − 1)(4)]
2

= 11[26 + 84]

= 1210

Thus, the required sum is 1210.

Page : 199 , Block Name : Miscellaneous Exercise

Q7 If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that
n
f (1) = 3 and ∑ f (x) = 120, find the value of n
x=1

Answer. It is given that,
f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)
f (1) = 3
Taking x = y = 1 in (1), we obtain
f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9
Similarly,
f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27
f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81
∴ f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the rst term and common ratio equal to
3.
n
a(r −1)
It is known that, S n =
r−1

It is given that, (\sum_{x=1}^{n} f(x)=120\)
n
3(3 −1)
∴ 120 =
3−1

3 n
⇒ 120 = (3 − 1)
2

n
⇒ 3 − 1 = 80

n 4
⇒ 3 = 81 = 3

∴ n = 4

Thus, the value of n is 4.

Page : 199 , Block Name : Miscellaneous Exercise

Q8 The sum of some terms of G.P. is 315 whose rst term and the common ratio are 5 and 2,

respectively.
Find the last term and the number of terms.

Answer. Let the sum of n terms of the G.P. be 315.
n
a(r −1)
It is known that, S n =
r−1

It is given that the rst term a is 5 and common ratio r is 2.
n
5(2 −1)
∴ 315 =
2−1

n
⇒ 2 − 1 = 63

n 6
⇒ 2 = 64 = (2)

⇒ n = 6

∴Last term of the G.P = 6th term = ar6 – 1 = (5)(2)5 = (5)(32) = 160
th 6−1 5
∴ LLast term of the G.P = 6 term = ar = (5)(2) = (5)

(32) = 160

Thus, the last term of the G.P. is 160.

Page : 199 , Block Name : Miscellaneous Exercise

Q9 The rst term of a G.P. is 1. The sum of the third term and fth term is 90.
Find the common ratio of G.P.

Answer. Let a and r be the rst term and the common ratio of the G.P. respectively.
∴a=1
2 2
a3 = ar = r

4 4
a5 = ar = r

2 4
∴ r + r = 90

4 2
⇒ r + r − 90 = 0
−1+√1+360 −1±√361 −1±19
2
⇒ r = = = = −10 or 9
2 2 2

∴ r = ±3 (Taking real roots)

Thus, the common ratio of the G.P. is ±3.

Page : 199 , Block Name : Miscellaneous Exercise

Q10 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order,
we obtain an arithmetic progression. Find the numbers.

Answer.
Let the three numbers in G.P. be a ar, and ar?

2
From the given condition, a + ar + ar = 56
2
⇒ a (1 + r + r ) = 56

56
⇒ a = 2
… (1)
1+r+r

2
a − 1, ar − 7, ar − 21 forms an A.P.

2
∴ (ar − 7) − (a − 1) = (ar − 21) − (ar − 7)

2
⇒ ar − a − 6 = ar − ar − 14

2
⇒ ar − 2ar + a = 8

2
⇒ ar − ar − ar + a = 8

2
⇒ a(r − 1) = 8 … (2)
56 2
⇒ 2
(r − 1) = 8
1+r+r

2
⇒ 7 + r + r

2 2
⇒ 7r − 14r + 7 − 1 + r + r

2 2
⇒ 7r − 14r + 7 − 1 − r − r = 0

2
⇒ 6r − 15r + 6 = 0
2
⇒ 6r − 12r − 3r + 6 = 0

⇒ 6r(r − 2) − 3(r − 2) = 0

⇒ (6r − 3)(r − 2) = 0

1
∴ r = 2,
2

When r = 2, a = 8

1
When r = , a = 32
2

Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.
When r = , the three numbers in G.P. are 32, 16, and 8.
1

2

Thus, in either case, the three required numbers are 8, 16, and 32.

Page : 199 , Block Name : Miscellaneous Exercise

Q11 A G.P. consists of an even number of terms.
If the sum of all the terms is 5 times the sum of terms occupying odd places, then nd its common
ratio.

Answer.
Let the G.P. be T1 , T2 , T3 , … T4 , … T2n .

Number of terms = 2n

According to the given condition,

T1 + T2 + T3 + … + T2n = 5 [T1 + T3 + … + T2n−1 ] = 0

⇒ T1 + T2 + T3 + … + T2n − 5 [T1 + T3 + … + T2n−1 ] = 0

⇒ T2 + T4 + … + T2n = 4 [T1 + T3 + … + T2n−1 ] = 0
2 3
Let the G.P. be a, ar, ar , ar , …
n n
ar(r −1) 4×a(r −1)
∴ =
r−1 r−1

⇒ ar = 4a

⇒ r = 4

Thus, the common ratio of the G.P. is 4.

Page : 199 , Block Name : Miscellaneous Exercise

Q12 The sum of the rst four terms of an A.P. is 56. The sum of the last four terms is 112.

If its rst term is 11, then nd the number of terms.

Answer. Let the A.P. be a, a + d, a + 2d, a + 3d, … a + (n – 2) d, a + (n – 1)d.
Sum of rst four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d
Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d]
+ [a + n – 1) d]
= 4a + (4n – 10) d
According to the given condition,
4a + 6d = 56
⇒ 4(11) + 6d = 56 [Since a = 11 (given)]
⇒ 6d = 12
⇒d=2
∴ 4a + (4n –10) d = 112
⇒ 4(11) + (4n – 10)2 = 112
⇒ (4n – 10)2 = 68
⇒ 4n – 10 = 34
⇒ 4n = 44
⇒ n = 11
Thus, the number of terms of the A.P. is 11.

Page : 199 , Block Name : Miscellaneous Exercise

Q13 If then show that a, b, c and d are in G.P.
a+bx b+cx c+dx
= = (x ≠ 0)
a−bx b−cx c−dx

Answer.
It is given that,

a+bx b+cx
=
a−bx b−cx

⇒ (a + bx)(b − cx) = (b + cx)(a − bx)

2 2 2 2
⇒ ab − acx + b x − bcx = ab − b x + acx − bcx

2
⇒ b x = 2acx

2
⇒ b = ac

b c
⇒ = … (1)
a b
b+cx c+dx
Also, =
b−cx c−dx

⇒ (b + cx)(c − dx) = (b − cx)(c + dx)

2 2 2 2
⇒ bc − bdx + c x − cdx = bc + bdx − c x − cdx

2
⇒ 2c x = 2bdx

2
⇒ c = bd

c d
⇒ = … (2)
d c

From (1) and (2), we obtain

b c d
= =
a b c

Thus, a, b, c, and d are in G.P.

Page : 199 , Block Name : Miscellaneous Exercise

Q14 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P.
Prove that P R = S 2 n n

Answer.
2 3 n−1
Let the G.P. be a, ar, ar , ar , … . ar …

According to the given information,
n
a(r −1)
S =
r−1

n 1+2+…+n−1
P = a × r
n−11
(n+1)
n
= a r 2 [∵ Sum of first n natural numbers is n ]
2

1 1 1
R = + + … +
n−1
a ar ar
n−1 n−2
r + r + …r + 1
=
n−1
ar
n
1 (r − 1) 1
n−1
= × [∵ 1, r, . . . . r f rom a G. P . ]
n−1
(r − 1) ar
n
r −1
= n−1
ar (r−1)

n n
(r −1)
2 n 2n n(n−1)
∴ P R = a r
n n(n−1) n
a r (r−1)

n n n
a (r − 1)
=
n
(r − 1)
n
n n
a(r − 1)
= [ ]
(r − 1)

n
= S

Hence, P R 2 n
= S
n

Page : 199 , Block Name : Miscellaneous Exercise

Q15 The p th
,q
th
andr
th
terms of an A.P. are a, b, c, respectively. Show that (q – r )a + (r – p )b + (p – q
)c = 0

Answer. Let t and d be the rst term and the common difference of the A.P. respectively.
The nth term of an A.P. is given by, an = t + (n – 1) d
Therefore,
ap = t + (p – 1) d = a … (1)
aq = t + (q – 1)d = b … (2)
ar = t + (r – 1) d = c … (3)
Subtracting equation (2) from (1), we obtain
(p – 1 – q + 1) d = a – b
⇒ (p – q) d = a – b
a−b
∴ d = … (4)
p−q

Subtracting equation (3) from (2), we obtain
(q – 1 – r + 1) d = b – c
⇒ (q – r) d = b – c

b−c
⇒ d = . . . (5)
q−r

Equating both the values of d obtained in (4) and (5), we obtain
a−b b−c
=
p−q q−r

⇒ (a − b)(q − r) = (b − c)(p − q)

⇒ aq − bq − ar + br = bp − cp + cq

⇒ bp − cp + cq − aq + ar − br = 0

⇒ (−aq + ar) + (bp − br) + (−cp + cq) = 0

⇒ −a(q − r) − b(r − p) − c(p − q) = 0

⇒ a(q − r) + b(r − p) + c(p − q) = 0

Thus, the given result is proved.

Page : 199 , Block Name : Miscellaneous Exercise

Q16 If a ( 1
+
1

c
),b(
1

c
+
1

a
),c(
1

a
+
1
) are in A.P., prove that a, b, c are in A.P.
b b

Answer.
1 1 1 1 1 1
It is given that a ( + ),b( + ) , c( + )
b c c a a b
are in A.P.

1 1 1 1 1 1 1 1
∴ b( + ) − a( + ) = c( + ) − b( + )
c a b c a b c a

b(a+c) a(b+c) c(a+b) b(a+c)
⇒ − = −
ac bc ab ac

2 2 2 2 2 2 2 2
b a+b c−a b−a c c a+c b−b a−b c
⇒ =
abc abc

2 2 2 2 2 2 2 2
⇒ b a − a b + b c − a c = c a − b a + c b − b c

2 2 2 2
⇒ ab(b − a) + c (b − a ) = a (c − b ) + bc(c − b)

⇒ (b − a)(ab + cb + ca) = (c − b)(c + b) + bc(c − b)

⇒ (b − a)(ab + c − b) = (c − b)(ac + ab + bc)

⇒ b − a = c − b

Thus, a, b, and c are in A.P.

Page : 199 , Block Name : Miscellaneous Exercise

Q17 If a, b, c, d are in G.P, prove that (a n n
+ b ) , (b
n n
+ c ) , (c
n
+ d
n
) are in G.P.

Answer. It is given that a, b, c,and d are in G.P.
2
∴ b = ac … (1)

2
c = bd … (2)

ad = bc … (3)

n n n n n n
It has to be proved that (a + b ) , (b + c ) , (c + d ) are in G.P. i.e.,

n n 2 n n n n
(b + c ) = (a + b ) (c + d )

Consider L.H.S.

n n 2 2n n n 2n
(b + c ) = b + 2b c + c
n n
2 n n 2
= (b ) + 2b c + (c )

n n n n
= (ac) + 2b c + (bd) [U sing(1) and (2)]

n n n n n n n n
= a c + b c + b c + b d
n n n n n n n n
= a c + b c + a d + b d [ Using (3)]

n n n n n n
= c (a + b ) + d (a + b )

n n n n
= (a + b ) (c + d )

= R. H. S.
n n 2 n n n n
∴ (b + c ) = (a + b ) (c + d )

n n n n n n
Thus, (a + b ) , (b + c ) , and (c + d ) are in G.P.

Page : 199 , Block Name : Miscellaneous Exercise

Q18
2 2
If a and b are the roots of x − 3x + p = 0 and c, d are roots of x − 12x + q = 0

where a, b, c, d form a G.P. Prove that (q + p) : (q − p) = 17 : 15.

Answer.
2
It is given that a and b are the roots of x − 3x + p = 0

∴ a + b = 3 and ab = p … (1)

2
Also, c and d are the roots of x − 12x + q = 0

∴ c + d = 12 and cd = q … . (2)

It is given that a, b, c, d are in G.P.
2 3
Let a = x, b = xr, c = xr , d = xr

From (1) and (2), we obtain

x + xr = 3

⇒ x(1 + r) = 3

2 3
xr + xr = 12

2
⇒ xΓ (1 + r) = 12

On dividing, we obtain
2
xr (1+r) 12
=
x(1+r) 3

2
⇒ r = 4

2
⇒ r = 4

⇒ r = ±2

3 3
When r = −2, x = = = 1
1−2 3

3 3
When r = −2, x = = = −3
1−2 −1

Case : I

When r = 2 and x = 1

2
ab = x r = 2

2 5
cd = x r = 32

q+p 32+2 34 17
∴ = = =
q−p 32−2 30 15

i.e., (q + p) : (q − p) = 17 : 15

Case : II
When r = −2, x = −3

2
ab = x r = −18

2 5
cd = x r = −288

q+p −288−18 −306 17
∴ = = =
q−p −288+18 −270 15

i.e., (q + p) : (q − p) = 17 : 15

Thus, in both the cases, we obtain (q + p): (q – p) = 17:15

Page : 199 , Block Name : Miscellaneous Exercise

Q19 The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n.
Show that a : b = (m + √m 2 2
− n ) : (m − √m
2 2
− n )

Answer. Let the two numbers be a and b.
a+b
A.M = and G.M. = √ab
2

According to the given condition,
a+b m
=
n
2√ab

2
(a+b) 2
m
⇒ =
2
4(ab) n

2
2 4abm
⇒ (a + b) = 2
n

2√abm
⇒ (a + b) =
n
2 2
Using this in the identity (a − b) = (a + b) − 4ab, we

obtain
2 2
2 4ab(m −n )
2 4abm
(a − b) = − 4ab =
2 2
n n

2 2
2√ab√m −n
⇒ (a − b) = … (2)
n

Adding (1) and (2), we obtain

2√ab
2 2
2a = (m + √m − n )
n

√ab
2 2
⇒ a = (m + √m − n )
n

Substituting the value of a in (1), we obtain

2√ab √ab
2 2
b = m − (m + √m − n )
n n

√ab √ab
= m − √ m2 − n 2
n n

√ab
(m − √m
2 2
= − n )
n

√ab
2 2
(m+√m −n ) (m+√m2 −n2 )
a √ab

∴ a : b = = =
b n 2 2
(m−√m −n )

2 2 2 2
Thus, a : b = (m + √m − n ) : (m − √m − n )

Page : 200 , Block Name : Miscellaneous Exercise

Q20 If a, b, c are in A.P.; b, c, d are in G.P. and 1

c
,
1

d
,
1

e
are in A.P.
prove that a, c, e are in G.P.

Answer. It is given that a, b, c are in A.P.
∴ b – a = c – b … (1)
It is given that b, c, d, are in G.P.
2
∴ c = bd … (2)

1 1 1
Also, , , are in A.P.
c d c

1 1 1 1
− = −
d c e d

2 1 1
= + … (3)
d c e

It has to be proved that a, c, e are in G.P. i.e., a = ae 2

From (1), we obtain
2b = a + c
a+c
⇒ b =
2

From (2), we obtain
2
c
d =
b

Substituting these values in (3), we obtain
2b 1 1
= +
c
2 c e

2(a+c) 1 1
⇒ = +
2c2 c e

a+c e+c
⇒ 2
=
c ce

a+c e+c
⇒ =
c c

⇒ (a + c)e = (e + c)c

2
⇒ ae + ce = ec + c

2
⇒ c = ae

Thus, a, c, and e are in G.P.

Page : 200 , Block Name : Miscellaneous Exercise

Q21 Find the sum of the following series up to n terms:
(i) 5 + 55 +555 + … (ii) .6 +. 66 +. 666+…

Answer. (i) 5 + 55 + 555 + …

Let Sn = 5 + 55 + 555 + ….. to n terms
5
= [9 + 99 + 999 + … to n terms ]
9

5 2 3
= [(10 − 1) + (10 − 1) + (10 − 1) + … to n terms ]
9

5 2 3
= [(10 + 10 + 10 + … n terms ) − (1 + 1 + … n terms )]
9
n
5 10 (10 − 1)
= [ − n]
9 10 − 1
n
5 10 (10 − 1)
= [ − n]
9 9

50 5n
n
= (10 − 1) −
81 9

(ii) .6 +.66 +. 666 +…
Let Sn = 06. + 0.66 + 0.666 + … to n terms
= 6[0.1 + 0.11 + 0.111 + … . to n terms ]

6
= [0.9 + 0.99 + 0.999 + … to n terms ]
9

6 1 1 1
= [(1 − ) + (1 − ) + (1 − ) + … ton terms ]
2 3
9 10 10 10

2 1 1 1
= [(1 + 1 + … n terms ) − (1 + + + … . terms )]
2
3 10 10 10
n
1
⎡ ⎛ 1 − ( ) ⎞⎤
2 1 10

= ⎢n − ⎜ ⎟⎥
⎢ ⎜ ⎟⎥
3 10 1
1 −
⎣ ⎝ 10 ⎠⎦

2 2 10
−n
= n − × (1 − 10 )
3 30 9

2 2
−n
= n − (1 − 10 )
3 27

Page : 200 , Block Name : Miscellaneous Exercise

Q22 Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms.

Answer.
The given series is 2 × 4 + 4 × 6 + 6 × 8 + … n terms

th 2
∴ n term = an = 2n × (2n + 2) = 4n + 4n

2
a20 = 4(20) + 4(20) = 4(400) + 80 = 1600 + 80 = 1680

th
Thus, the 20 term of the series is 1680.

Page : 200 , Block Name : Miscellaneous Exercise

Q23 Find the sum of the rst n terms of the series: 3+ 7 +13 +21 +31 +…

Answer.

The given series is 3 + 7 + 13 + 21 + 31 + …

S = 3 + 7 + 13 + 21 + 31 + … + an−1 + an

S = 3 + 7 + 13 + 21 + … + an−2 + an−1 + an

On subtracting both the equations, we obtain
S − S = [3 + (7 + 13 + 21 + 31 + … + an−1 + an )] − [(3 + 7+13 + 21 + 31 + … + an−1 ) + an ]

S − S = 3 + [(7 − 3) + (13 − 7) + (21 − 13) + … + (an − an−1 )] − an

0 = 3 + [4 + 6 + 8 + … (n − 1) terms ] − an

an = 3 + [4 + 6 + 8 + … (n − 1) terms ]

n − 1
⇒ an = 3 + ( ) [2 × 4 + (n − 1 − 1)2]
2

n − 1
= 3 + ( ) [8 + (n − 2)2]
2

(n − 1)
= 3 + (2n + 4)
2

= 3 + (n − 1)(n + 2)

2
= 3 + (n + n − 2)

2
= n + n + 1
n n n n

2
∴ ∑ ak = ∑ k + ∑ k + ∑ 1

k=1 k=1 k=1 k=1

n(n + 1)(2n + 1) n(n + 1)
= + + n
6 2

(n + 1)(2n + 1) + 3(n + 1) + 6
= n[ ]
6

2
2n + 3n + 1 + 3n + 3 + 6
= n[ ]
6

2
2n + 3n + 10
= n[ ]
6

n
2
= (n + 3n + 5)
3

Page : 200 , Block Name : Miscellaneous Exercise

Q24
If S1 , S2 , S3 are the sum of first n natural numbers, their squares and their

2
cubes, respectively, show that 9S = S3 (1 + 8S1 )
2

Answer. From the given information,

n(n + 1)
S1 =
2
2 2
n (n + 1)
S3 =
4
2 2
n (n + 1) 8n(n + 1)
Here, S3 (1 + 8S1 ) = [1 + ]
4 2

2 2
n (n + 1)
2
= [1 + 4n + 4n]
4
2 2
n (n + 1)
2
= (2n + 1)
4
2
[n(n+1)(2n+1)]
= … (1)
4

2
2 [n(n+1)(2n+1)]
Also, 9S2 = 9
4

9 2
= [n(n + 1)(2n + 1)]
36

2
[n(n+1)(2n+1)]
= … (2)
4

Thus, from (1) and (2), we obtain 9S
2
= S3 (1 + 8S1 )
2

Page : 200 , Block Name : Miscellaneous Exercise

Q25 Find the sum of the following series up to n terms:
3 3 3 3 3 3
1 1 +2 1 +2 +3
+ + + …
1 1+3 1+3+5

Answer. The nth term of the given series is
2
n(n+1)

3 3 3 3 [ ]
1 +2 +3 +…+n 2

=
1+3+5+…+(2n−1) 1+3+5+…+(2n−1)

Here,1,3,5,.... (2n-1) is an A.P. with rst term a, last term (2n-1) and number of terms as n
n 2
∴ 1 + 3 + 5 + … + (2n − 1) = [2 × 1 + (n − 1)2] = n
2

2 2 2
n (n+1) (n+1) 1 1 1
2
∴ an = 2
= = n + n +
4n 4 4 2 4

n n 1 2 1 1
∴ Sn = ∑ aK = ∑ ( K + K + )
k=1 k=1 4 2 4

1 n(n + 1)(2n + 1) 1 n(n + 1) 1
= + + n
4 6 2 2 4

n[(n + 1)(2n + 1) + 6(n + 1) + 6]
=
24
2
n [2n + 3n + 1 + 6n + 6 + 6]
=
24
2
n (2n + 9n + 13)
=
24

Page : 200 , Block Name : Miscellaneous Exercise

2 2 2
1×2 +2×3 +…+n×(n+1)
Q26 Show that 2 2 2
=
3n+5

3n+1
.
1 ×2+2 ×3+…+n ×(n+1)

Answer.
th 2 3 2
n term of the numerator = n(n + 1) = n + 2n + n

th 2 3 2
n term of the denominator = n (n + 1) = n + n
2 2 n n 3 2
2
1×2 +2×3 +…+n×(n+1) ∑ ak ∑ (K +2K +K)
k=1 k=1

2 2
= n = n . . . (1)
3 2
1 ×2+2 ×3+…+n2 ×(n+1) ∑k=1 ak ∑k=1 (K +K )

n 3 2
Here, ∑ (K + 2K + K)
K=1

2 2
n (n+1) 2n(n+1)(2n+1) n(n+1)
= + +
4 6 2

n(n+1)[n(n+1) 2
= + (2n + 1) + 1]
2 3

n(n+1) 2
3n +3n+8n+4+6
= [ ]
2 6

n(n + 1)
2
= [3n + 1ln + 10]
12
n(n + 1)
2
= [3n + 6n + 5n + 10]
12

n(n + 1)
= [3n(n + 2) + 5(n + 2)]
12

n(n + 1)(n + 2)(3n + 5)
= … (2)
12
2 2
n 3 2 n (n+1) n(n+1)(2n+1)
Also, ∑ (K + K ) = +
k=1 4 6

n(n+1) n(n+1) 2n+1
= [ + ]
2 2 3

n(n+1) 2
3n +3n+4n+2
= [ ]
2 6

n(n + 1)
2
= [3n + 7n + 2]
12
n(n + 1)
2
= [3n + 6n + n + 2]
12

n(n + 1)
= [3n(n + 2) + 1(n + 2)]
12

n(n + 1)(n + 2)(3n + 1)
= … (3)
12

From (1), (2), and (3), we obtain
n(n+1)(n+2)(3n+5)
2 2 2
1×2 +2×3 +…+n×(n+1) 12

2 2
=
n(n+1)(n+2)(3n+1)
1 ×2+2 ×3+…+n2 ×(n+1)
12

n(n+1)(n+2)(3n+5) 3n+5
= =
n(n+1)(n+2)(3n+1) 3n+1

Thus, the given result is proved.

Page : 200 , Block Name : Miscellaneous Exercise

Q27 A farmer buys a used tractor for Rs 12000.
He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest
on the unpaid amount.
How much will the tractor cost him?

Answer. It is given that the farmer pays Rs 6000 in cash.
Therefore, unpaid amount = Rs 12000 – Rs 6000 = Rs 6000
According to the given condition, the interest paid annually is
12% of 6000, 12% of 5500, 12% of 5000, …, 12% of 500
Thus, total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500
= 12% of (6000 + 5500 + 5000 + … + 500)
= 12% of (500 + 1000 + 1500 + … + 6000)
Now, the series 500, 1000, 1500 … 6000 is an A.P. with both the rst term and common difference equal
to 500.
Let the number of terms of the A.P. be n.
∴ 6000 = 500 + (n – 1) 500
⇒ 1 + (n – 1) = 12
⇒ n = 12
∴Sum of the A.P
12
= [2(500) + (12 − 1)(500)] = 6[1000 + 5500] = 6(6500) = 39000
2

Thus, total interest to be paid = 12% of (500 + 1000 + 1500 + … + 6000)
= 12% of 39000 = Rs 4680
Thus, cost of tractor = (Rs 12000 + Rs 4680) = Rs 16680

Page : 200 , Block Name : Miscellaneous Exercise

Q28 Shamshad Ali buys a scooter for Rs 22000.
He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest
on the unpaid amount.
How much will the scooter cost him?

Answer. It is given that Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.
∴Unpaid amount = Rs 22000 – Rs 4000 = Rs 18000
According to the given condition, the interest paid annually is
10% of 18000, 10% of 17000, 10% of 16000 … 10% of 1000
Thus, total interest to be paid = 10% of 18000 + 10% of 17000 + 10% of 16000 + … + 10% of 1000
= 10% of (18000 + 17000 + 16000 + … + 1000)
= 10% of (1000 + 2000 + 3000 + … + 18000)
Here, 1000, 2000, 3000 … 18000 forms an A.P. with rst term and common difference both equal to
1000.
Let the number of terms be n.
∴ 18000 = 1000 + (n – 1) (1000)
⇒ n = 18
18
∴ 1000 + 2000 + … . +18000 = [2(1000) + (18 − 1)(1000)]
2

= 9[2000 + 17000]

= 171000

∴ Total interest paid = 10% of (18000 + 17000 + 16000 + … + 1000)
= 10% of Rs 171000 = Rs 17100
∴Cost of scooter = Rs 22000 + Rs 17100 = Rs 39100

Page : 200 , Block Name : Miscellaneous Exercise

Q29 A person writes a letter to four of his friends.
He asks each one of them to copy the letter and mail to four different persons with instruction that
they move the chain similarly.
Assuming that the chain is not broken and that it costs 50 paise to mail one letter.
Find the amount spent on the postage when 8th set of letter is Mailed.

Answer. The numbers of letters mailed forms a G.P.: 4, 4 , … 4 2 8

First term = 4
Common ratio = 4
Number of terms = 8
It is known that the sum of n terms of a G.P. is given by
n
a(r −1)
Sn =
r−1
s
4(4 −1) 4(65536−1) 4(65535)
∴ Sg = = = = 4(21845) = 87380
4−1 3 3

It is given that the cost to mail one letter is 50 paisa.
∴Cost of mailing 87380 letters = Rs 87380 × == Rs 43690
50

100

Thus, the amount spent when 8th set of letter is mailed is Rs 43690.

Page : 200 , Block Name : Miscellaneous Exercise

Q30 A man deposited Rs 10000 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.

Answer. It is given that the man deposited Rs 10000 in a bank at the rate of 5% simple interest
annually.
∴ Interest in rst year = 5
× Rs10000 = Rs500
100

∴Amount in 15th year = Rs 10000 + 500 + 500 + … + 500

14 times

= Rs 10000 + 14 × Rs 500
= Rs 10000 + Rs 7000
= Rs 17000
Amount after 20 years = Rs10000 + 500 + 500 + … . +500

20 times

= Rs 10000 + 20 × Rs 500
= Rs 10000 + Rs 10000

Page : 200 , Block Name : Miscellaneous Exercise

Q31 A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate
each year by 20%.
Find the estimated value at the end of 5 years.

Answer. Cost of machine = Rs 15625
Machine depreciates by 20% every year.

Therefore, its value after every year is 80% of the original cost i.e., 4

5
of the original cost.
∴ Value at the end of 5 years = 15625 × 4

5
×
4

5
× ….×
4

5
= 5 × 1024 = 5120
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5 times