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

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3.

• Sequences and Series Class 11 Ex 9.1
• Sequences and Series Class 11 Ex 9.2
• Sequences and Series Class 11 Ex 9.4

Board	CBSE
Textbook	NCERT
Class	Class 11
Subject	Maths
Chapter	Chapter 9
Chapter Name	Sequences and Series
Exercise	Ex 9.3
Number of Questions Solved	32
Category	NCERT Solutions

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

Question 1.
Find the 20^th and n^th terms of the G.P. , ….
Solution:

Question 2.
Find the 12^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.
Solution:
We have, a_s = 192, r = 2

Question 3.
The 5^th, 8^th and 11th terms of a G.P. are p, q and s, respectively. Show that q² = ps.
Solution:
We are given

Question 4.
The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7^th term.
Solution:
We have a= -3, a₄ = (a₂)²

Question 5.
Which term of the following sequences:


Solution:

Question 6.
For what values of x, the numbers , x, are in G.P.?
Solution:

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:

Question 7.
0.14, 0.015, 0.0015, …. 20 items.
Solution:
In the given G.P.

Question 8.
, …. n terms
Solution:
In the given G.P.

Question 9.
1, -a, a²,- a³ … n terms (if a ≠ -1)
Solution:
In the given G.P.. a = 1, r = -a

Question 10.
x³, x⁵, ⁷, ….. n terms (if ≠±1).
Solution:
In the given G.P., a = x³, r = x²

Question 11.
Evaluate
Solution:

Question 12.
The sum of first three terms of a G.P. is and 10 their product is 1. Find the common ratio and the terms.
Solution:
Let the first three terms of G.P. be ,a,ar, where a is the first term and r is the common ratio.

Question 13.
How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?
Solution:
Let n be the number of terms we needed. Here a = 3, r = 3, S_n = 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.
Solution:
Let a₁, a₂, a₃, a₄, a₅, a₆ be the first six terms of the G.P.

Question 15.
Given a G.P. with a = 729 and 7^th term 64, determine S₇.
Solution:
Let a be the first term and the common ratio be r.

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.
Solution:
Let a₁ a₂ be first two terms and a₃ a₅ be third and fifth terms respectively.
According to question

Question 17.
If the 4^th, 10^th and 16^th 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, then according to question

Question 18.
Find the sum to n terms of the sequence, 8, 88, 888, 8888 ………
Solution:
This is not a G.P., however we can relate it to a G.P. by writing the terms as S_n= 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,
Solution:
On multiplying the corresponding terms of sequences, we get 256, 128, 64, 32 and 16, which forms a G.P. of 5 terms

Question 20.
Show that the products of the corresponding terms of the sequences a, ar, ar², ………… ar^n-1 and A, AR, AR², …….. , AR^n-1 form a G.P., and find the common ratio.
Solution:
On multiplying the corresponding terms, we get aA, aArR, aAr²R²,…… aAr^n-1R^n-1. We can see that this new sequence is G.P. with first term aA & the common ratio rR.

Question 21.
Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4^th by 18.
Solution:
Let the four numbers forming a G.P. be a, ar, ar², ar³
According to question,

Question 22.
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 A be the first term and R be the common ratio, then according to question

Question 23.
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P² – (ab)ⁿ.
Solution:
Let r be the common ratio of the given G.P., then b = n^th term = ar^n-1

Question 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
Solution:
Let the G.P. be a, ar, ar², ……
Sum of first n terms = a + ar + ……. + ar^n-1

Question 25.
If a, b,c and d are in G.P., show that
(a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²
Solution:
We have a, b, c, d are in G.P.
Let r be a common ratio, then

Question 26.
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Solution:
Let G₁, G₂ be two numbers between 3 and 81 such that 3, G₁ G₂,81 is a G.P.

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

Question 28.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio .
Solution:
Let a and b be the two numbers such that a + b = 6

Question 29.
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are .
Solution:
Let a and b be the numbers such that A, G are A.M. and G.M. respectively between them.

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?
Solution:
There were 30 bacteria present in the culture originally and it doubles every hour. So, the number of bacteria at the end of successive hours form the G.P. i.e., 30, 60, 120, 240, …….

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?
Solution:
We have, Principal value = Rs. 500 Interest rate = 10% annually

Question 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 α & β be the roots of a quadratic equation such that A.M. & G.M. of α, β are 8 and 5 respectively.

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