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

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

Chapter 9: Sequences and Series in Class 11 Maths is part of the CBSE Syllabus for 2024-25. The NCERT Solutions for this chapter help students master problems related to sequences and series. Subject experts have solved all the questions in this exercise. Exercise 9.3 of NCERT Solutions for Class 11 Maths Chapter 9 covers:

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• Geometric Progression (G.P.)
• The general term of a G.P.
• Sum to n terms of a G.P.
• Geometric Mean (G.M.)
• Relationship between A.M. and G.M.

These solutions are prepared by subject matter experts at Infinity Learn, providing detailed methods for solving problems. By understanding the concepts in these solutions, students can clear their doubts and excel in their board exams.

## NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series Exercise 9.3 – PDF Download

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

### NCERT Solutions for Class 11 Maths Chapter 9 Sequence and series Exercise 9.3

1. The 4th term of a G.P. is the 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, a4 = ar3 = (–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.

2. The sum of the 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.

3. 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–rbr–pcp–q

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

= Aqr + rp + pq × R (prprq + r) + (rq r + ppq) + (prpqr + q)

= A0 × R0

= 1

Hence proved.

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

Solution:

Given, a, b, c, d 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 + 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.

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

Therefore,

(a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2

5. 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, G1, G2, 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.

6. 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.

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