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:

• 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

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

a_n = ar^n–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 p^th, q^th and r^th terms of a G.P. are a, b and c, respectively. Prove that a^q-r b^r-p c^p-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

AR^p–1 = a

AR^q–1 = b

AR^r–1 = c

Then,

a^q–rb^r–pc^p–q

= A^q–r × R^{(p–1) (q–r)} × A^r–p × R^{(q–1) (r–p)} × A^p–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

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

bc = ad … (1)

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 1^st term as 3 and the 4^th 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 2^nd hour, 4^th hour and n^th 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 2^nd hour will be 120.

And, a₅ = ar⁴ = (30) (2)⁴ = 480

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

a_{n +1} = arⁿ = (30) 2ⁿ

Therefore, the number of bacteria at the end of n^th hour will be 30(2)ⁿ.