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

#### Page No 192:

#### Question 1:

Find the 20

^{th}and*n*^{th}terms of the G.P.#### Answer:

The given G.P. is

Here,

*a*= First term =*r*= Common ratio =

#### Question 2:

Find the 12

^{th}term of a G.P. whose 8^{th}term is 192 and the common ratio is 2.#### Answer:

Common ratio,

*r*= 2
Let

*a*be the first term of the G.P.
∴

*a*_{8}=*ar*^{8–1}=*ar*^{7}
⇒

*ar*^{7}= 192*a*(2)

^{7}= 192

*a*(2)

^{7}= (2)

^{6}(3)

#### 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*.#### Answer:

Let

*a*be the first term and*r*be the common ratio of the G.P.
According to the given condition,

*a*

_{5}=

*a*

*r*

^{5–1 }=

*a*r

^{4}=

*p*… (1)

*a*

_{8 }=

*a*

*r*

^{8–1 }=

*a*

*r*

^{7}=

*q*… (2)

*a*

_{11}= a

*r*

^{11–1 }=

*a*

*r*

^{10}=

*s*… (3)

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

Dividing equation (3) by (2), we obtain

Equating the values of

*r*^{3}obtained in (4) and (5), we obtain
Thus, the given result is proved.

#### 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.#### Answer:

Let

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

*a*= –3
It is known that,

*a**=*_{n}*ar*^{n}^{–1}
∴

*a*_{4 }=*ar*^{3}= (–3)*r*^{3}*a*

_{2}=

*a r*

^{1}= (–3)

*r*

According to the given condition,

(–3)

*r*^{3}= [(–3)*r*]^{2}
⇒ –3

*r*^{3}= 9*r*^{2}
⇒

*r*= –3*a*

_{7}=

*a*

*r*

^{7–1 }=

*a*

*r*

^{6}= (–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)

#### Answer:

**(a)**The given sequence is

Here,

*a*= 2 and*r*=
Let the

*n*^{th}term of the given sequence be 128.
Thus, the 13

^{th}term of the given sequence is 128.**(b)**The given sequence is

Here,

Let the

*n*^{th}term of the given sequence be 729.
Thus, the 12

^{th}term of the given sequence is 729.**(c)**The given sequence is

Here,

Let the

*n*^{th}term of the given sequence be .
Thus, the 9

^{th}term of the given sequence is .#### Question 6:

For what values of

*x,*the numbers are in G.P?#### Answer:

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 …

#### Answer:

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#### Answer:

The given G.P. is

Here,

#### Question 9:

Find the sum to

*n*terms in the geometric progression#### Answer:

The given G.P. is

Here, first term =

*a*_{1}= 1
Common ratio =

*r*= –*a*#### Question 10:

Find the sum to

*n*terms in the geometric progression#### Answer:

The given G.P. is

Here,

*a*=*x*^{3}and*r*=*x*^{2}#### Question 11:

Evaluate

#### Answer:

The terms of this sequence 3, 3

^{2}, 3^{3}, … 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.

#### Answer:

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

From (2), we obtain

*a*

^{3}= 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, 3

^{2}, 3^{3}, … are needed to give the sum 120?#### Answer:

The given G.P. is 3, 3

^{2}, 3^{3}, …
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.#### Answer:

Let the G.P. be

*a*,*ar*,*ar*^{2},*ar*^{3}, …
According to the given condition,

*a*+

*ar*+

*ar*

^{2}= 16 and

*ar*

^{3 }+

*ar*

^{4}+

*ar*

^{5 }= 128

⇒

*a*(1 +*r*+*r*^{2}) = 16 … (1)*ar*

^{3}(1 +

*r*+

*r*

^{2}) = 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 7^{th}term 64, determine S_{7}.#### Answer:

*a*= 729

*a*

_{7}= 64

Let

*r*be the common ratio of the G.P.
It is known that,

*a*_{n}=*a r*^{n}^{–1}*a*

_{7}=

*ar*

^{7–1}= (729)

*r*

^{6}

⇒ 64 = 729

*r*^{6}
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.

#### Answer:

Let

*a*be the first term and*r*be the common ratio of the G.P.
According to the given conditions,

*a*

_{5}= 4 ×

*a*

_{3}

*ar*

^{4}= 4

*ar*

^{2}

⇒

*r*^{2}= 4
∴

*r*= ± 2
From (1), we obtain

Thus, the required G.P. is

4, –8, 16, –32, …

#### 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.#### Answer:

Let

*a*be the first term and*r*be the common ratio of the G.P.
According to the given condition,

*a*

_{4}=

*a*

*r*

^{3}=

*x*… (1)

*a*

_{10}=

*a*

*r*

^{9}=

*y*… (2)

*a*

_{16}

^{ }=

*a r*

^{15 }=

*z*… (3)

Dividing (2) by (1), we obtain

Dividing (3) by (2), we obtain

∴

Thus,

*x*,*y*,*z*are in G. P.#### Page No 193:

#### Question 18:

Find the sum to

*n*terms of the sequence, 8, 88, 888, 8888…#### Answer:

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

*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, .

#### Answer:

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.

#### Answer:

It has to be proved that the sequence,

*aA*,*arAR*,*ar*^{2}*AR*^{2}, …*ar*^{n}^{–1}*AR*^{n}^{–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 4

^{th}by 18.#### Answer:

Let

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

_{1}=

*a*,

*a*

_{2}=

*ar*,

*a*

_{3}=

*ar*

^{2},

*a*

_{4}=

*ar*

^{3}

By the given condition,

*a*

_{3}=

*a*

_{1}+ 9

⇒

*ar*^{2}=*a*+ 9 … (1)*a*

_{2}=

*a*

_{4}+ 18

⇒

*ar*=*ar*^{3}+ 18 … (2)
From (1) and (2), we obtain

*a*(

*r*

^{2}– 1) = 9 … (3)

*ar*(1–

*r*

^{2}) = 18 … (4)

Dividing (4) by (3), we obtain

Substituting the value of

*r*in (1), we obtain
4

*a*=*a*+ 9
⇒ 3

*a*= 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#### Answer:

Let

*A*be the first term and*R*be the common ratio of the G.P.
According to the given information,

*AR*

^{p}

^{–1 }=

*a*

*AR*

^{q}

^{–1 }=

*b*

*AR*

^{r}

^{–1 }=

*c*

*a*

^{q–r}

^{ }

*b*

^{r–p}

*c*

^{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}^{)}
=

*A*^{0}×*R*^{0}
= 1

Thus, the given result is proved.

#### Question 23:

If the first and the

*n*^{th}term of a G.P. are*a*ad*b*, respectively, and if*P*is the product of*n*terms, prove that*P*^{2}= (*ab*)^{n}.#### Answer:

The first term of the G.P is

*a*and the last term is*b*.
Therefore, the G.P. is

*a*,*ar*,*ar*^{2},*ar*^{3}, …*ar*^{n}^{–1}, where*r*is the common ratio.*b*=

*ar*

^{n}

^{–1}… (1)

*P*= Product of

*n*terms

= (

*a*) (*ar*) (*ar*^{2}) … (*ar*^{n}^{–1})
= (

*a*×*a*×…*a*) (*r*×*r*^{2}× …*r*^{n}^{–1})
=

*a*^{n}*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 .#### Answer:

Let

*a*be the first term and*r*be the common ratio of the G.P.
Since there are

*n*terms from (*n*+1)^{th}to (2*n*)^{th}term,
Sum of terms from(

*n*+ 1)^{th}to (2*n*)^{th}term*a*

^{n }

^{+1}=

*ar*

^{n + 1}

^{– 1}=

*ar*

^{n}

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 (2*n*)^{th }term is^{ .}#### Question 25:

If

*a, b, c*and*d*are in G.P. show that .#### Answer:

*a*,

*b*,

*c*,

*d*are in G.P.

Therefore,

*bc*=

*ad*… (1)

*b*

^{2}=

*ac*… (2)

*c*

^{2}=

*bd*… (3)

It has to be proved that,

(

*a*^{2}+*b*^{2}+*c*^{2}) (*b*^{2}+*c*^{2}+*d*^{2}) = (*ab*+*bc*–*cd*)^{2}
R.H.S.

= (

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

*ab*+*ad*+*cd*)^{2}[Using (1)]
= [

*ab*+*d*(*a*+*c*)]^{2}
=

*a*^{2}*b*^{2}+ 2*abd*(*a*+*c*) +*d*^{2}(*a*+*c*)^{2}
=

*a*^{2}*b*^{2}+2*a*^{2}*bd*+ 2*acbd*+*d*^{2}(*a*^{2}+ 2*ac*+*c*^{2})
=

*a*^{2}*b*^{2}+ 2*a*^{2}*c*^{2}+ 2*b*^{2}*c*^{2}+*d*^{2}*a*^{2}+ 2*d*^{2}*b*^{2}+*d*^{2}*c*^{2}[Using (1) and (2)]
=

*a*^{2}*b*^{2}+*a*^{2}*c*^{2}+*a*^{2}*c*^{2}+*b*^{2}*c*^{2 }+*b*^{2}*c*^{2}+*d*^{2}*a*^{2}+*d*^{2}*b*^{2}+*d*^{2}*b*^{2}+*d*^{2}*c*^{2}
=

*a*^{2}*b*^{2}+*a*^{2}*c*^{2}+*a*^{2}*d*^{2 }+*b*^{2 }×*b*^{2}+*b*^{2}*c*^{2}+*b*^{2}*d*^{2}+*c*^{2}*b*^{2}+*c*^{2 }×*c*^{2}+*c*^{2}*d*^{2}
[Using (2) and (3) and rearranging terms]

=

*a*^{2}(*b*^{2}+*c*^{2}+*d*^{2}) +*b*^{2}(*b*^{2}+*c*^{2}+*d*^{2}) +*c*^{2}(*b*^{2}+*c*^{2}+*d*^{2})
= (

*a*^{2}+*b*^{2}+*c*^{2}) (*b*^{2}+*c*^{2}+*d*^{2})
= 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.

#### Answer:

Let

*G*_{1}and*G*_{2}be two numbers between 3 and 81 such that the series, 3,*G*_{1},*G*_{2}, 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}
⇒

*r*^{3}= 27
∴

*r*= 3 (Taking real roots only)
For

*r*= 3,*G*

_{1}=

*ar*= (3) (3) = 9

*G*

_{2}=

*ar*

^{2}= (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*.#### Answer:

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.

#### Answer:

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.

#### Answer:

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*= 2

*A*… (3)

*ab*=

*G*

^{2}… (4)

Substituting the value of

*a*and*b*from (3) and (4) in the identity (*a*–*b*)^{2}= (*a*+*b*)^{2}– 4*ab*, we obtain
(

*a*–*b*)^{2}= 4*A*^{2}– 4*G*^{2}= 4 (*A*^{2}–*G*^{2})
(

*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 2

^{nd}hour, 4^{th}hour and*n*^{th}hour?#### Answer:

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
∴

*a*_{3}=*ar*^{2}= (30) (2)^{2}= 120
Therefore, the number of bacteria at the end of 2

^{nd}hour will be 120.*a*

_{5}=

*ar*

^{4}= (30) (2)

^{4}= 480

The number of bacteria at the end of 4

^{th}hour will be 480.*a*

_{n}

_{ +1 }=

*ar*

*= (30) 2*

^{n}^{n}

Thus, number of bacteria at the end of

*n*^{th}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?

#### Answer:

The amount deposited in the bank is Rs 500.

At the end of first year, amount = = Rs 500 (1.1)

At the end of 2

^{nd}year, amount = Rs 500 (1.1) (1.1)
At the end of 3

^{rd}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.

#### Answer:

Let the root of the quadratic equation be

*a*and*b*.
According to the given condition,

The quadratic equation is given by,

*x*

^{2}–

*x*(Sum of roots) + (Product of roots) = 0

*x*

^{2}–

*x*(

*a*+

*b*) + (

*ab*) = 0

*x*

^{2}– 16

*x*+ 25 = 0 [Using (1) and (2)]

Thus, the required quadratic equation is

*x*^{2}– 16*x*+ 25 = 0_{}

^{}