**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 the part of NCERT Solutions for Class 11 Maths.
Here you can find the NCERT Solutions for Class 11 Maths Chapter 9 Sequences
and Series Ex 9.3.

**Ex 9.3
Class 11 Maths Question 1.**

Find the 20^{th}and n

^{th}terms of the G.P. 5/2, 5/4, 5/8, ….

**Solution:**

**Ex 9.3 Class 11
Maths 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_{8} = 192 and r = 2.

**Ex 9.3 Class 11
Maths Question 3.**

The 5^{th}, 8

^{th}and 11

^{th}terms of a G.P. are p, q and s, respectively. Show that q

^{2}= ps.

**Solution:**

We are given that

**Ex 9.3 Class 11 Maths Question
4.**

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

^{th}term.

**Solution:**

Given, a = -3, a_{4} = (a_{2})^{2}

**Ex 9.3 Class 11 Maths Question 5.**

Which term
of the following sequences:**Solution:**

**Ex 9.3 Class 11 Maths Question 6.**

For what values of x, the numbers −2/7, x, −7/2 are in G.P.?

**Solution:**

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

**Ex 9.3 Class 11
Maths Question 7.**

0.15, 0.015, 0.0015, …. 20 terms.

**Solution:**

In the given G.P., we have,

**Ex 9.3 Class 11 Maths Question 8.**

√7, √21, 3√7, …. n terms

**Solution:**

In the given G.P., we have,

**Ex 9.3 Class 11
Maths Question 9.**

1, -a, a^{2}, -a

^{3}, … n terms (if a ≠ -1)

**Solution:**

In the given G.P., a = 1, r = -a.

**Ex 9.3 Class 11
Maths Question 10.**

x^{3}, x

^{5}, x

^{7}, ….. n terms (if ≠ ±1).

**Solution:**

In the given G.P., a = x^{3}, r = x^{2}

**Ex 9.3 Class 11
Maths Question 11.**

Evaluate

**Solution:**

**Ex 9.3 Class 11 Maths Question 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 the first three terms of the G.P. be a/r, a, ar, where a is the first term and r is the
common ratio.

**Ex 9.3 Class 11 Maths Question
13.**

How many terms of G.P. 3, 3^{2}, 3

^{3}, … are needed to give the sum 120?

**Solution:**

Let n be the number of terms we needed. Here a = 3, r = 3 and S_{n} =
120

**Ex 9.3 Class 11
Maths 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 a_{1}, a_{2}, a_{3}, a_{4}, a_{5},
a_{6} be the first six terms of the G.P.

**Ex 9.3 Class 11
Maths Question 15.**

Given a G.P. with a = 729 and 7^{th}term 64, determine S

_{7}.

**Solution:**

Let a be the first term and the common ratio be r.

**Ex 9.3 Class 11
Maths 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 a_{1}, a_{2} be first two terms and a_{3}, a_{5} be
third and fifth terms, respectively.

According to the question,

**Ex 9.3 Class 11
Maths 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 the
question,

**Ex 9.3 Class 11
Maths 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

**Ex 9.3 Class 11 Maths 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, ½.

**Solution:**

By multiplying the corresponding terms of the given sequences, we get

256, 128, 64, 32 and 16, which forms a G.P. of 5
terms

**Ex 9.3 Class 11
Maths Question 20.**

Show that the products of the corresponding terms of the sequences a, ar, ar^{2}, ………… ar

^{n-1}and A, AR, AR

^{2}, …….., AR

^{n-1}form a G.P., and find the common ratio.

**Solution:**

By multiplying the corresponding terms of the two sequences, we get

aA, aArR, aAr^{2}R^{2},
……, aAr^{n-1}R^{n-1}.

We can see that this new
sequence forms a G.P. with first term aA & the common ratio rR.

**Ex 9.3 Class 11
Maths 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 the G.P. be a, ar, ar^{2}, ar^{3}

According to the question,

**Ex 9.3 Class 11
Maths 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 of the given G.P., then
according to the question,

**Ex 9.3 Class 11
Maths Question 23.**

If the first and the n^{th}term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P

^{2}= (ab)

^{n}.

**Solution:**

Let r be the common ratio of the given G.P., then b = n^{th} term
= ar^{n-1}

**Ex 9.3 Class 11 Maths 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 1/r

^{n}.

**Solution:**

Let the G.P. be a, ar, ar^{2}, ……

The sum of first n terms = a + ar + ……. + ar^{n-1}

**Ex 9.3 Class 11
Maths Question 25.**

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

^{2}+ b

^{2}+ c

^{2}) (b

^{2}+ c

^{2}+ d

^{2}) = (ab + bc + cd)

^{2}

**Solution:**

Given, a, b, c and d are in G.P.

Let r be the common ratio, then

**Ex 9.3 Class 11
Maths Question 26.**

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

**Solution:**

Let G_{1} and G_{2} be two numbers between 3 and 81 such
that 3, G_{1}, G_{2}, 81 is a G.P.

**Ex 9.3 Class 11 Maths Question 27.**

Find the value of n so that (a^{n}

^{+}

^{1}+ b

^{n}

^{+}

^{1})/(a

^{n}+ b

^{n}) may be the geometric mean between a and b.

**Solution:**

**Ex 9.3 Class 11 Maths Question 28.**

The sum of two numbers is 6 times their geometric mean, show that numbers are
in the ratio (3 + 2√2) : (3 − 2√2).

**Solution:**

Let a and b be the two numbers such that a + b = 6 √ab

**Ex 9.3 Class 11 Maths 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 two numbers such that A and G are A.M. and G.M.,
respectively between them.

**Ex 9.3 Class 11
Maths 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 2^{nd}hour, 4

^{th}hour and n

^{th}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, …….

**Ex 9.3 Class 11
Maths 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:**

Given, Principal value = Rs. 500 and Interest rate = 10% compounded annually

**Ex 9.3 Class 11
Maths 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 Î± and Î² be the roots of the quadratic equation such that A.M. and G.M. of Î±
and Î² are 8 and 5, respectively.

**You can also like these:**

**NCERT Solutions for Maths Class 9**

**NCERT Solutions for Maths Class 10**