Class 11 Maths NCERT Solutions for Chapter 9 Sequences and Series Exercise 9.3
Sequences and Series Exercise 9.3 Solutions
1. Find the 20th and nth terms of the G.P. 5/2, 5/4, 5/8, ......
Solution
The given G.P. is 5/2, 5/4, 5/8 , .....
Here, a = First term = 5/2
r = Common ratio = 
2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Solution
Common ratio, r = 2
Let a be the first term of the G.P.
∴ a8 = ar 8–1 = ar7
⇒ ar7 = 192
a(2)7 = 192
⇒ a(2)7 = (2)6 (3) 
3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.
Solution
Let a be the first 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 
Dividing equation (3) by (2), we obtain 
Equating the values of r3 obtained in (4) and (5), we obtain
q/p = s/q
⇒ q2 = ps
Thus, the given result is proved.
4. The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.
Solution
Let a be the first term and r be the common ratio of the G.P.
∴ a = –3
It is known that, an = arn–1
∴a4 = ar3 = (–3) r3
a2 = a r1 = (–3) r
According to the given condition,
(–3) r3 = [(–3) r]2
⇒ –3r3 = 9 r2
⇒ r = –3
a7 = a r 7–1 = a r6 = (–3) (–3)6 = – (3)7 = –2187
Thus, the seventh term of the G.P. is –2187.
5. Which term of the following sequences:
(a) 2, 2√2, 4, ..... is 128 ?
(b) √3, 3, 3√3, .... is 729?
(c) 1/3, 1/9, 1/27, .... is 1/19683 ?
Solution
(a) The given sequence is 2, 2√2, 4, .....
Here, a = 2 and r = 2√2/2 = √2
Let the nth term of the given sequence be 128. 
⇒ 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.
an = arn-1
∴ arn-1 = 729
⇒ (√3)(√3)n-1 = 729 
⇒ n = 12
Thus, the 12th term of the given sequence is 729.
(c) The given sequence is 1/3, 1/9, 1/27, ......
Here, a = 1/3 and r = 1/9 ÷ 1/3 = 1/3
Let the nth term of the given sequence be 1/19683. 
Thus, the 9th term of the given sequence is 1/19683.
6. For what values of x, the numbers 2/7 , x, -7/2 are in G.P?
Solution
⇒ x = √1
⇒ x = ± 1
Thus, for x = ± 1, the given numbers will be in G.P.
Here, a = 0.15 and r = 0.015/0.15 = 0.1
Here, a = √7
Here, first term = a1 = 1
Common ratio = r = -a
Here, a = x3 and r = x2
a/r + a + ar = 39/10 ...(1)
(a/r)(a) (ar) = 1 ...(2)
From (2), we obtain
a3 = 1
⇒ a = 1 (Considering real roots only)
Substituting a = 1 in equation (1), we obtain
⇒ 10 + 10r + 10r2 - 39r = 0
⇒ 10r2 - 29r + 10 = 0
⇒ 10r2 - 25r - 4r + 10 = 0
⇒ 5r(2r - 5) - 2(2r - 5) = 0
⇒ (5r - 2)(2r - 5) = 0
⇒ r = 2/5 or 5/2
Thus, the three terms of G.P. are 5/2, 1, and 2/5.
The given G.P. is 3, 32, 33, …
Let n terms of this G.P. be required to obtain the sum as 120.
⇒ 3n - 1 = 80
⇒ 3n = 81
⇒ 3n = 34
n = 4
Thus, four terms of the given G.P. are required to obtain the sum as 120.
14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Solution
Let the G.P. be a, ar, ar2, ar3, …
According to the given condition,
a + ar + ar2 = 16 and ar3 + ar4 + ar5 = 128
⇒ a (1 + r + r2) = 16 …(1)
ar3(1 + r + r2) = 128 …(2)
Dividing equation (2) by (1), we obtain
⇒ r3 = 8
∴ r = 2
Substituting r = 2 in (1), we obtain
a(1 + 2 +4) = 16
⇒ a(7) = 16
⇒ a = 16/7 
15. Given a G.P. with a = 729 and 7th term 64, determine S7.
Solution
a = 729
a7 = 64
Let r be the common ratio of the G.P.
It is known that, an = a rn–1
a7 = ar7–1 = (729)r6
⇒ 64 = 729 r6
= (3)7 - (2)7
= 2187 - 128
= 2059
According to the given conditions, 
a5 = 4 × a3
ar4 = 4ar2
⇒ r2 = 4
∴ r = ± 2
From (1), we obtain 
⇒ a = 4
Thus, the required G.P. is
-4/3, -8/3, -16/3, ..... or 4, -8, 16, -32, .....
17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Solution
Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a4 = a r3 = x …(1)
a10 = a r9 = y …(2)
a16= a r15 = z …(3)
Dividing (2) by (1), we obtain 
18. Find the sum to n terms of the sequence, 8, 88, 888, 8888…
Solution
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 
19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2
Solution
Required sum = 2 × 128 + 4 × 32 + 8 × 8 + 16 × 2 + 32 × 1/2
= 64[4 + 2 + 1 + 1/2 + 1/22].
Here, 4, 2, 1, 1/2, 1/22 is a G.P.
First term, a = 4
Common ratio, r = 1/2
It is known that, 
Required sum = 64(31/4) = 16 × 31 = 496
20. Show that the products of the corresponding terms of the sequences a, ar, ar2, …arn – 1 and A, AR, AR2, …ARn – 1 form a G.P, and find the common ratio
Solution
It has to be proved that the sequence, aA, arAR, ar2AR2, …arn–1ARn–1, forms a G.P.
Second term/First term = arAR/aA = rR
Third term/Second term = ar2 AR2 /ar AR = rR
Thus, the above sequence forms a G.P. and the common ratio is rR.
21. Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Solution
Let a be the first term and r be the common ratio of the G.P.
a1 = a, a2 = ar, a3 = ar2, a4 = ar3
By the given condition,
a3 = a1 + 9
⇒ ar2 = a + 9 …(1)
a2 = a4 + 18
⇒ ar = ar3 + 18 …(2)
From (1) and (2), we obtain
a(r2 – 1) = 9 …(3)
ar (1– r2) = 18 …(4)
Dividing (4) by (3), we obtain
⇒ -r = 2
⇒ r = -2
Substituting the value of r in (1), we obtain
4a = a + 9
⇒ 3a = 9
∴ a = 3
Thus, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3(–2)3 i.e., 3¸–6, 12, and –24.
22. If the pth , qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq-r br-p cp-q = 1
Solution
Let A be the first term and R be the common ratio of the G.P.
According to the given information,
ARp–1 = a
ARq–1 = b
ARr–1 = c
aq–rbr–pcp–q
= Aq–r × R(p–1) (q–r) × Ar–p × R(q–1) (r-p) × Ap–q × R(r –1)(p–q)
= Aq – r + r – p + p – q × R (pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)
= A0 × R0
= 1
Thus, the given result is proved.
23. If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.
Solution
The first term of the G.P is a and the last term is b.
Therefore, the G.P. is a, ar, ar2, ar3, … arn–1, where r is the common ratio.
b = arn–1 … (1)
P = Product of n terms
= (a) (ar) (ar2) … (arn–1)
= (a × a ×…a) (r × r2 × …rn–1)
= anr 1 + 2 +…(n–1) … (2)
Here, 1, 2, …(n – 1) is an A.P. 
Thus, the given result is proved.
24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n +1)th to (2n)th term is 1/rn .
Solution
Let a be the first term and r be the common ratio of the G.P.
Sum of first n terms a(l - rn )/(1 - r)
Since there are n terms from (n + 1)th to (2n)th term,
Sum of terms from (n + 1)th to (2n)th term = 
an+1 = arn+1-1 = arn
Thus, required ratio = 
Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is 1/rn.
25. If a, b, c and d are in G.P. show that (a2 + b2 + c2 ) (b2 + c2 + d2 ) =(ab + bc + cd)2 .
Solution
a, b, c, d are in G.P.
Therefore,
bc = ad …(1)
b2 = ac …(2)
c2 = bd …(3)
It has to be proved that,
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2
R.H.S.
= (ab + bc + cd)2
= (ab + ad + cd)2 [Using (1)]
= [ab + d (a + c)]2
= a2b2 + 2abd (a + c) + d2 (a + c)2
= a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2)
= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 [Using (1) and (2)]
= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2
= a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2
[Using (2) and (3) and rearranging terms]
= a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2)
= (a2 + b2 + c2) (b2 + c2 + d2)
= L.H.S.
∴ L.H.S. = R.H.S.
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2
26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Solution
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 first term and r be the common ratio of the G.P.
∴81 = (3) (r)3
⇒ r3 = 27
∴ r = 3 (Taking real roots only)
For r = 3,
G1 = ar = (3) (3) = 9
G2 = ar2 = (3) (3)2 = 27
Thus, the required two numbers are 9 and 27.
27. Find the value of n so that (an+1 + bn+1)/(an + bn) may be the geometric mean between a and b.
Solution
G. M. of a and b is √ab.
By the given condition, 
Squaring both sides, we obtain 
28. The sum of two numbers is 6 time their geometric mean, show that numbers are in the ratio (3 + 2√2) : (3 - 2√2) .
Solution
Let the two numbers be a and b.
G.M. = √ab
According to the given condition,
a + b = 6√ab ...(1)
⇒ (a + b)2 = 36(ab)
Also,
(a - b)2 = (a + b)2 - 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 
Thus, the required ratio is (3 + 2√2):(3 - 2√2).
29. 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)
Solution
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.
∴ AM = A = (a + b)/2 ...(1)
GM = G = √(ab) ...(2)
From (1) and (2), we obtain
a + b = 2A …(3)
ab = G2 …(4)
Substituting the value of a and b from (3) and (4) in the identity (a – b)2 = (a + b)2 – 4ab, we obtain
(a – b)2 = 4A2 – 4G2 = 4 (A2–G2)
(a – b)2 = 4 (A + G) (A – G)
From (3) and (5), we obtain 
30. 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 nthhour?
Solution
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
∴ a3 = ar2 = (30) (2)2 = 120
Therefore, the number of bacteria at the end of 2nd hour will be 120.
a5 = ar4 = (30)(2)4 = 480
The number of bacteria at the end of 4th hour will be 480.
an +1 = arn = (30) 2n
Thus, number of bacteria at the end of nth hour will be 30(2)n.
31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Solution
The amount deposited in the bank is Rs 500.
At the end of first 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.
32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Solution
Let the root of the quadratic equation be a and b.
According to the given condition,
A.M. = (a + b)/2 = 8 ⇒ a + b = 16 ...(1)
G.M. = √ab = 5 ⇒ ab = 25 ...(2)
The quadratic equation is given by,
x2– x (Sum of roots) + (Product of roots) = 0
x2 – x (a + b) + (ab) = 0
x2 – 16x + 25 = 0 [Using (1) and (2)]
Thus, the required quadratic equation is x2 – 16x + 25 = 0