 # NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series Exercise 9.3

NCERT exercise solutions of this chapter are helpful for the students to improve their hold on the problems related to sequences and series. All the questions of this exercise have been solved by subject experts. Exercise 9.3 of NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series is based on the following topics:

1. Geometric Progression (G. P.)
1. The general term of a G.P
2. Sum to n terms of a G.P
3. Geometric Mean (G.M.)
2. Relationship Between A.M. and G.M.

These solutions are prepared by subject matter experts at BYJU’S, describing the complete method of solving problems. By understanding the concepts used in NCERT Solutions for Class 11 Maths, students will be able to clear all their doubts related to the chapters of Class 11.

## Download PDF of NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series Exercise 9.3                 ### Access Other Exercise Solutions of Class 11 Maths Chapter 9- Sequences and Series

Exercise 9.1 Solutions 14 Questions

Exercise 9.2 Solutions 18 Questions

Exercise 9.4 Solutions 10 Questions

Miscellaneous Exercise On Chapter 9 Solutions 32 Questions

#### Access Solutions for Class 11 Maths Chapter 9.3 Exercise

1. Find the 20th and nthterms of the G.P. 5/2, 5/4, 5/8, ………

Solution:

Given G.P. is 5/2, 5/4, 5/8, ………

Here, a = First term = 5/2

r = Common ratio = (5/4)/(5/2) = ½

Thus, the 20th term and nth term 2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

Solution:

Given,

The common ratio of the G.P., r = 2

And, let a be the first term of the G.P.

Now,

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 pq and s, respectively. Show that q2 = ps.

Solution:

Let’s take a to be the first term and r to be the common ratio of the G.P.

Then according to the question, we have

a5 = a r5–1 a r4 = p … (i)

aa r8–1 a r7 = q … (ii)

a11 = a r11–1 a r10 = … (iii)

Dividing equation (ii) by (i), we get 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’s consider a to be the first term and r to be the common ratio of the G.P.

Given, a = –3

And we know that,

an = arn–1

So, aar3 = (–3) r3

a2 = a r1 = (–3) r

Then from the question, we have

(–3) r3 = [(–3) r]2

⇒ –3r3 = 9 r2

⇒ r = –3

a7 = a r 7–1 a r6 = (–3) (–3)6 = – (3)7 = –2187

Therefore, 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, 2, 2√2, 4,…

We have,

a = 2 and r = 2√2/2 = √2

Taking the nth term of this sequence as 128, we have Therefore, the 13th term of the given sequence is 128.

(ii) Given sequence, √3, 3, 3√3,…

We have,

a = √3 and r = 3/√3 = √3

Taking the nth term of this sequence to be 729, we have Therefore, the 12th term of the given sequence is 729.

(iii) Given sequence, 1/3, 1/9, 1/27, …

a = 1/3 and r = (1/9)/(1/3) = 1/3

Taking the nth term of this sequence to be 1/19683, we have Therefore, 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:

The given numbers are -2/7, x, -7/2.

Common ratio = x/(-2/7) = -7x/2

Also, common ratio = (-7/2)/x = -7/2x Therefore, for x = ± 1, the given numbers will be in G.P.

7. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …

Solution:

Given G.P., 0.15, 0.015, 0.00015, …

Here, a = 0.15 and r = 0.015/0.15 = 0.1 8. Find the sum to n terms in the geometric progression √7, √21, 3√7, ….

Solution:

The given G.P is √7, √21, 3√7, ….

Here,

a = √7 and 9. Find the sum to n terms in the geometric progression 1, -a, a2, -a3 …. (if a ≠ -1)

Solution:

The given G.P. is 1, -a, a2, -a3 ….

Here, the first term = a1 = 1

And the common ratio = r = – a

We know that, 10. Find the sum to n terms in the geometric progression x3, x5, x7, … (if x ≠ ±1 )

Solution:

Given G.P. is x3, x5, x7, …

Here, we have a = x3 and r = x5/x3 = x2 11. Evaluate: Solution: 12. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

Solution:

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

a/r + a + ar = 39/10 …… (1)

(a/r) (a) (ar) = 1 …….. (2)

From (2), we have

a3 = 1

Hence, a = 1 [Considering real roots only]

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

1/r + 1 + r = 39/10

(1 + r + r2)/r = 39/10

10 + 10r + 10r2 = 39r

10r2 – 29r + 10 = 0

10r2 – 25r – 4r + 10 = 0

5r(2r – 5) – 2(2r – 5) = 0

(5r – 2) (2r – 5) = 0

Thus,

r = 2/5 or 5/2

Therefore, the three terms of the G.P. are 5/2, 1 and 2/5.

13. How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?

Solution:

Given G.P. is 3, 32, 33, …

Let’s consider that n terms of this G.P. be required to obtain the sum of 120.

We know that, Here, a = 3 and r = 3 Equating the exponents we get, n = 4

Therefore, 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’s assume the G.P. to be aarar2ar3, …

Then according to the question, we have

ar + ar2 = 16 and arar4 + ar= 128

a (1 + r + r2) = 16 … (1) and,

ar3(1 + r + r2) = 128 … (2)

Dividing equation (2) by (1), we get r3 = 8

r = 2

Now, using r = 2 in (1), we get

a (1 + 2 + 4) = 16

a (7) = 16

a = 16/7

Now, the sum of terms is given as 15. Given a G.P. with a = 729 and 7th term 64, determine S7.

Solution:

Given,

a = 729 and a7 = 64

Let r be the common ratio of the G.P.

Then we know that, an = a rn–1

a7 = ar7–1 = (729)r6

⇒ 64 = 729 r6

r6 = 64/729

r6 = (2/3)6

r = 2/3

And, we know that 16. Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term.

Solution:

Consider a to be the first term and r to be the common ratio of the G.P.

Given, S2 = -4

Then, from the question we have And,

a5 = 4 x a3

ar4 = 4ar2

r2 = 4

r = ± 2

Using the value of r in (1), we have Therefore, 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)

On dividing (2) by (1), we get 18. Find the sum to n terms of the sequence, 8, 88, 888, 8888…

Solution:

Given sequence: 8, 88, 888, 8888…

This sequence is not a G.P.

But, 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:

The required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x ½

= 64[4 + 2 + 1 + ½ + 1/22]

Now, it’s seen that

4, 2, 1, ½, 1/22 is a G.P.

With first term, a = 4

Common ratio, r =1/2

We know,  Therefore, the 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:

To be proved: The sequence, aAarARar2AR2, …arn–1ARn–1, forms a G.P.

Now, we have Therefore, 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:

Consider a to be the first term and r to be the common ratio of the G.P.

Then,

a1 = aa2 = ara3 = ar2a4 = ar3

From the question, we have

a3 = a1 + 9

ar2 = a + 9 … (i)

a2 = a4 + 18

ar ar3 + 18 … (ii)

So, from (1) and (2), we get

a(r2 – 1) = 9 … (iii)

ar (1– r2) = 18 … (iv)

Now, dividing (4) by (3), we get -r = 2

r = -2

On substituting the value of r in (i), we get

4a + 9

3a = 9

∴ a = 3

Therefore, 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’s take A to be the first term and R to be the common ratio of the G.P.

Then according to the question, we have

ARp–1 a

ARq–1 b

ARr–1 c

Then,

aq–r br–p cp–q

Aq× R(p–1) (q–r) × Arp × R(q–1) (rp) × Apq × R(–1)(pq)

Aq – r + r – p + p – q × R (pr – pr – q + r) + (rq – r p – pq) + (pr – p – qr + q)

A0 × R0

= 1

Hence 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:

Given, the first term of the G.P is a and the last term is b.

Thus,

The G.P. is aarar2ar3, … arn–1, where r is the common ratio.

Then,

b = arn–1  … (1)

P = Product of n terms

= (a) (ar) (ar2) … (arn–1)

= (a × a ×…a) (r × r2 × …rn–1)

an r 1 + 2 +…(n–1)  … (2)

Here, 1, 2, …(n – 1) is an A.P. And, the product of n terms P is given by, 24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from .

Solution:

Let a be the first term and be the common ratio of the G.P. Since there are n terms from (n +1)th to (2n)th term,

Sum of terms from(n + 1)th to (2n)th term a +1 = ar n + 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 .

25. If a, b, c and d are in G.P. show that (a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2.

Solution:

Given, abcd are in G.P.

So, we have

b2 = ac  … (2)

c2 = bd  … (3)

Taking the R.H.S. we have

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 + b2cb2c2 + d2a2 + d2b2 + d2b2 + d2c2

a2b2 + a2c2 + a2db× b2 + b2c2 + b2d2 + c2b2 + c× c2 + c2d2

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

a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2c2 + d2)

= (a2 + b2 + c2) (b2 + c2 + d2)

= L.H.S.

Thus, L.H.S. = R.H.S.

Therefore,

(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’s assume G1 and G2 to be two numbers between 3 and 81 such that the series 3, G1G2, 81 forms a G.P.

And let a be the first term and r be the common ratio of the G.P.

Now, we have the 1st term as 3 and the 4th term as 81.

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

Therefore, the two numbers which can be inserted between 3 and 81 so that the resulting sequence becomes a G.P are 9 and 27.

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

Solution:

We know that,

The G. M. of a and b is given by √ab.

Then from the question, we have By squaring both sides, we get Performing cross multiplication after expanding, we get, 28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio .

Solution:

Consider the two numbers be a and b.

Then, G.M. = √ab.

From the question, we have 29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are .

Solution:

Given that A and G are A.M. and G.M. between two positive numbers.

And, let these two positive numbers be a and b. 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 nth hour?

Solution:

Given, the number of bacteria doubles every hour. Hence, the number of bacteria after every hour will form a G.P.

Here we have, a = 30 and r = 2

So, a3 = ar2 = (30) (2)2 = 120

Thus, the number of bacteria at the end of 2nd hour will be 120.

And, a5 = ar4 = (30) (2)4 = 480

The number of bacteria at the end of 4th hour will be 480.

an +1 arn = (30) 2n

Therefore, the 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:

Given,

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….

Therefore,

The 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’s consider the roots of the quadratic equation to be a and b.

Then, we have We know that,

A quadratic equation can be formed as,

x2 – x (Sum of roots) + (Product of roots) = 0

x2 – x (a + b) + (ab) = 0

x2 – 16x + 25 = 0 [Using (1) and (2)]

Therefore, the required quadratic equation is x2 – 16x + 25 = 0