# NCERT solution class 11 chapter 9 Sequences and Series exercise 9.3 mathematics

## EXERCISE 9.3

#### Question 1:

Find the 20th and nthterms of the G.P.

The given G.P. is

Here, a = First term =

r = Common ratio =

#### Question 2:

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

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)

#### Question 3:

The 5th, 8th and 11th terms of a G.P. are pq and s, respectively. Show that q2 = ps.

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)

aa r8–1 a r7 = q … (2)

a11 = a r11–1 a r10 = … (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

Thus, the given result is proved.

#### Question 4:

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

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

aar3 = (–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.

#### Question 5:

Which term of the following sequences:

(a)  (b)  (c)

(a) The given sequence is

Here, a = 2 and r =

Let the nth term of the given sequence be 128.

Thus, the 13th term of the given sequence is 128.

(b) The given sequence is

Here,

Let the nth term of the given sequence be 729.

Thus, the 12th term of the given sequence is 729.

(c) The given sequence is

Here,

Let the nth term of the given sequence be .

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

#### Question 6:

For what values of x, the numbers are in G.P?

The given numbers are .

Common ratio

Also, common ratio =

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

#### Question 7:

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

The given G.P. is 0.15, 0.015, 0.00015, …

Here, a = 0.15 and

#### Question 8:

Find the sum to n terms in the geometric progression

The given G.P. is

Here,

#### Question 9:

Find the sum to n terms in the geometric progression

The given G.P. is

Here, first term = a1 = 1

Common ratio = r = – a

#### Question 10:

Find the sum to n terms in the geometric progression

The given G.P. is

Here, a = x3 and r = x2

#### Question 11:

Evaluate

The terms of this sequence 3, 32, 33, … forms a G.P.

Substituting this value in equation (1), we obtain

#### Question 12:

The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.

Let be the first three terms of the G.P.

From (2), we obtain

a3 = 1

⇒ a = 1 (Considering real roots only)

Substituting a = 1 in equation (1), we obtain

Thus, the three terms of G.P. are .

#### Question 13:

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

The given G.P. is 3, 32, 33, …

Let n terms of this G.P. be required to obtain the sum as 120.

Here, a = 3 and r = 3

∴ n = 4

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

#### Question 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.

Let the G.P. be aarar2ar3, …

According to the given condition,

ar + ar2 = 16 and arar4 + ar= 128

⇒ a (1 + r + r2) = 16 … (1)

ar3(1 + r + r2) = 128 … (2)

Dividing equation (2) by (1), we obtain

Substituting r = 2 in (1), we obtain

a (1 + 2 + 4) = 16

⇒ a (7) = 16

#### Question 15:

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

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

Also, it is known that,

#### Question 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.

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

According to the given conditions,

a5 = 4 × a3

ar4 = 4ar2

⇒ r2 = 4

∴ = ± 2

From (1), we obtain

Thus, the required G.P. is

4, –8, 16, –32, …

#### Question 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.

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

Dividing (3) by (2), we obtain

∴

Thus, xyz are in G. P.

#### Question 18:

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

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

#### Question 19:

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

Required sum =

Here, 4, 2, 1, is a G.P.

First term, a = 4

Common ratio, r =

It is known that,

∴Required sum =

#### Question 20:

Show that the products of the corresponding terms of the sequences  form a G.P, and find the common ratio.

It has to be proved that the sequence, aAarARar2AR2, …arn–1ARn–1, forms a G.P.

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

#### Question 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.

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

a1 = aa2 = ara3 = ar2a4 = 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

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

4a + 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.

#### Question 22:

If the terms of a G.P. are a, b and c, respectively. Prove that

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–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

Thus, the given result is proved.

#### Question 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.

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

Therefore, the G.P. is aarar2ar3, … 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)

an r 1 + 2 +…(n–1) … (2)

Here, 1, 2, …(n – 1) is an A.P.

∴1 + 2 + ……….+ (n – 1)

Thus, the given result is proved.

#### Question 24:

Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from .

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 .

#### Question 25:

If a, b, c and d are in G.P. show that .

abcd are in G.P.

Therefore,

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 + 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.

∴ L.H.S. = R.H.S.

#### Question 26:

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

Let G1 and G2 be two numbers between 3 and 81 such that the series, 3, G1G2, 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.

#### Question 27:

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

G. M. of a and b is .

By the given condition,

Squaring both sides, we obtain

#### Question 28:

The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio.

Let the two numbers be a and b.

G.M. =

According to the given condition,

Also,

Adding (1) and (2), we obtain

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

Thus, the required ratio is.

#### Question 29:

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are.

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.

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 (A2G2)

(a – b)2 = 4 (A + G) (A – G)

From (3) and (5), we obtain

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

Thus, the two numbers are.

#### Question 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?

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.

#### Question 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?

The amount deposited in the bank is Rs 500.

At the end of first year, amount = = 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

#### Question 32:

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

Let the root of the quadratic equation be a and b.

According to the given condition,

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

error: Content is protected !!