# Geometric Progression Concise Solutions Chapter-11 ICSE Maths

Geometric Progression Concise Solutions Chapter-11 for ICSE Maths Class 10th . Solutions of Exercise – 11 (A), Exercise –11 (B), Exercise – 11(C), Exercise –11 (D), and Additional Questions for Concise Selina Maths of ICSE Board Class 10th. Concise Solutions Geometric Progression Chapter – 11for ICSE Maths Class 10 is available here. Therefore All Solutions of Concise Selina of Chapter 11 Geometric Progression has been solved according instruction given by council. This is the  Solutions of Chapter-11 Geometric Progression for ICSE Class 10th. ICSE Maths text book of Concise is In series of famous ICSE writer in maths. Concise is most famous among students.

## Geometric Progression Concise Solutions Chapter-11 for ICSE Maths Class 10th

The Solutions of Concise Mathematics Chapter 11 Geometric Progression for ICSE Class 10 have been solved.  Hence Experience teachers Solved Chapter-11 Geometric Progression to help students of class 10th ICSE board. Therefore the ICSE Class 10th Maths Solutions of Concise Selina Publishers for Chapter-11 Geometric Progression helpful on  various topics which are prescribed in most ICSE Maths textbooks

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Exercise – 11 (A)Exercise – 11 (B) ,   Exercise – 11 (C) , Exercise – 11 (D) ,

### How to Solve Concise Maths Selina Publications Chapter-11 Geometric Progression

Note:- Before viewing Solutions of Chapter-11 Geometric Progression of Concise Selina Maths read the Chapter-11 Carefully then solve all example of your text book.  Chapter-11 Geometric Progression is main Chapter in ICSE board

### Exercise-11(A),Geometric Progression Concise Solutions

#### Question 1.

Find, which of the following sequence form a G.P. :
(i) 8, 24, 72, 216, ……
(ii) $\\ \frac { 1 }{ 8 }$,$\\ \frac { 1 }{ 24 }$,$\\ \frac { 1 }{ 72 }$,$\\ \frac { 1 }{ 216 }$
(iii) 9, 12, 16, 24,…..

(i) 8, 24, 72, 216,……
Here, a = 8

#### Question 2

Find the 9th term of the series :
1, 4, 16, 64,…….

In G.P. 1, 4, 16, 64,….
Here first term (a) = 1
and common ratio (r) = $\\ \frac { 4 }{ 1 }$ = 4,
T9 = arn – 1 = 1 x 49 – 1 = 1 x 48 = 48
= 4 x 4 x 4 x 4 x 4 x 4 x 4 x 4
= 65536

#### Question 3

Find the seventh term of the G.P. :
1 , √3, 3, 3√3…….

G.P. is 1 , √3, 3, 3√3
Here first term (a) = 1

#### Question 4

Find the 8th term of the sequence :
$\\ \frac { 3 }{ 4 }$,$1 \frac { 1 }{ 2 }$,3……

G.P. = $\\ \frac { 3 }{ 4 }$,$1 \frac { 1 }{ 2 }$,3…….
$\\ \frac { 3 }{ 4 }$,$\\ \frac { 3 }{ 2 }$,3…….

#### Question 5

Find the 10th term of the G.P. :
12, 4, $1 \frac { 1 }{ 3 }$,……

G.P. = 12, 4, $1 \frac { 1 }{ 3 }$,……..
= 12, 4, $\\ \frac { 4 }{ 3 }$,…..

#### Question 6

Find the nth term of the series :
1, 2, 4, 8 …….

1, 2, 4, 8,……
Here, a = 1,r = $\\ \frac { 2 }{ 1 }$ = 2

#### Question 7

Find the next three terms of the sequence :
√5, 5, 5√5…..

√5, 5, 5√5……
Here a = √5 and r = $\frac { 5 }{ \surd 5 }$ = √5

#### Question 8

Find the sixth term of the series :
22, 23, 24,….

22, 23, 24,……
Here, a = 22, r = 23 ÷ 22 = 23 – 2 = 21 = 2

#### Question  9

Find the seventh term of the G.P. :
√3 + 1, 1, $\frac { \surd 3-1 }{ 2 }$,…….

√3 + 1, 1, $\frac { \surd 3-1 }{ 2 }$,…….

#### Question 10

Find the G.P. whose first term is 64 and next term is 32.

First term of a G.P. (a) = 64
and second term (ar) = 32
G.P. will be 64, 32, 16, 8, 4, 2, 1,…….

#### Question 11

Find the next three terms of the series:
$\\ \frac { 2 }{ 27 }$,$\\ \frac { 2 }{ 9 }$,$\\ \frac { 2 }{ 3 }$,…..

G.P. is $\\ \frac { 2 }{ 27 }$,$\\ \frac { 2 }{ 9 }$,$\\ \frac { 2 }{ 3 }$,…..
a = $\\ \frac { 2 }{ 27 }$

#### Question 12

Find the next two terms of the series
2 – 6 + 18 – 54…..

G.P. is 2 – 6 + 18 – 54 +………
Here a = 2 and r = $\\ \frac { -6 }{ 2 }$ = – 3
Next two terms will be
– 54 x ( – 3) = + 162
162 x ( – 3) = – 486
Next two terms are 162 – 486

### Geometric Progression Concise Solutions Exercise-11(B) for ICSE Maths Class 10th

#### Question 1

.Which term of the G.P. :
– 10, $\frac { 5 }{ \surd 3 } ,-\frac { 5 }{ 6 } ,....-\frac { 5 }{ 72 } ?$

– 10, $\frac { 5 }{ \surd 3 } ,-\frac { 5 }{ 6 } ,....$
Here a = – 10
r = $\frac { 5 }{ \surd 3 } \div \left( -10 \right)$

n – 1 = 4
=> n = 4 + 1 = 5
It is 5th term

#### Question 2.

The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.

In a G.P.
T5 = ar4 = 81
T2 = ar = 24

#### Question 3.

Fourth and seventh terms of a G.P. $\\ \frac { 1 }{ 18 }$ are $- \frac { 1 }{ 486 }$ respectively. Find the GP.

In a G.P.

#### Question 4

.If the first and the third terms of a G.P. are 2 and 8 respectively, find its second term

In a G.P.
T1 = 2, and T3 = 8
=>a = 2 and ar² = 8
Dividing, we get
r² = $\\ \frac { 8 }{ 2 }$ = 4 = (2)²
r = 2
Second term = ar = 2 x 2 = 4

#### Question 5.

The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term.

Let a be first term and r be common ratio, then
T3 = ar2
and T8 = ar7
T3 x T8 = ar2 x ar7
= 243

#### Question 6.

Find the geometric progression with 4th term = 54 and 7th term = 1458.

In a G.P.
T4 = 54 and T7 = 1458
Let a be the first term and r be the common
ratio, then

#### Question 7.

Second term of a geometric progression is 6 and its fifth term is 9 times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.

In a G.P.
T2 = 6,
T5 = 9 x T3
Let a be the first term and r be the common ratio

#### Question 8.

The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.

In a G.P.
T4= 10,
T7 = 80 and l = 2560
Let a be the first term and r be the common ratio. Therefore

#### Question 9.

If the 4th and 9th terms of a G.P. are 54 and 13122 respectively, find the GP. Also, find its general term.

In a G.P.
T4 = 54 and T9 = 13122
Let a be the first term and r be the common ratio

#### Question 10.

The fifth, eight and eleventh terms of a geometric progression are p, q and r respectively. Show that : q² = pr.

In a G.P.
T5 = p,
T8 = q and T11 = r
To show that q² = pr
Let a be the first term and r be the common ratio, therefore
ar4 = p, ar7 = q and ar10 = r
Squaring the ar7 = q

### Geometric Progression Exercise-11(C)Concise Solutions

#### Question 1.

Find the seventh term from the end of the series :
√2, 2, 2√2,……32

√2, 2, 2√2,……32
Here a = √2
r = $\frac { 2 }{ \surd 2 } =\surd 2$
and l =32

#### Question 2.

Find the third term from the end of the GP.
$\frac { 2 }{ 27 } ,\frac { 2 }{ 9 } ,\frac { 2 }{ 3 } ,....162$

G.P is $\frac { 2 }{ 27 } ,\frac { 2 }{ 9 } ,\frac { 2 }{ 3 } ,....162$
a = $\\ \frac { 2 }{ 27 }$
r = $\frac { 2 }{ 9 } \div \frac { 2 }{ 27 }$
$\frac { 2 }{ 9 } \times \frac { 27 }{ 2 }$
= 3
l = 162

#### Question 3.

For the G.P. $\frac { 1 }{ 27 } ,\frac { 1 }{ 9 } ,\frac { 1 }{ 3 } .....81$
find the product of fourth term from the beginning and the fourth term from the end.

$\frac { 1 }{ 27 } ,\frac { 1 }{ 9 } ,\frac { 1 }{ 3 } .....81$
a = $\\ \frac { 2 }{ 27 }$
r = $\frac { 1 }{ 9 } \div \frac { 1 }{ 27 }$
$\frac { 1 }{ 9 } \times \frac { 27 }{ 1 }$
= 3

#### Question 4.

If for a G.P., pth, qth and rth terms are a, b and c respectively ;
prove that :
{q – r) log a + (r – p) log b + (p – q) log c = 0

In a G.P
Tp = a,
and Tq = b,
so Tr = c

#### Question 5.

If a, b and c in G.P., prove that : log an, log bn and log cn are in A.P.

a, b, c are in G.P.
Let A and R be the first term and common ratio respectively.
Therefore,
a = A
b = AR
c = AR2
log a = log A
and log b = log AR = log A + log R
log c = log AR2 = log A + 2log R
hence log a, log b and log c are in A.P.
If 2log b = log a + log c
then 2[logA + logR] = log A + log A + 2log R
If 2log A + 2log R = 2log A + 2log R
which is true.
Hence log a, log b and log c are in A.P.

#### Question 6.

If each term of a G.P. is raised to the power x, show that the resulting sequence is also a G.P.

Let a, b, c are in G.P.
Then b2 = ac …(i)
Now ax, bx + cx will be in G.P. if (bx)2 = ax.cx
=> (bx)2 = ax.cx
=>(b2)= (ac)x
Hence ax, bx, cx are in G.P. (∴ b2 = ac)
Hence proved.

#### Question 7.

If a, b and c are in A.P. a, x, b are in G.P. whereas b, y and c are also in G.P. Show that : x2, b2, y2 are in A.P.

2 b = a + c _(i)
a, x, b are in G.P.
x2 = ab _(ii)
and b, y, c in G.P.
y2 = bc _(iii)
Now x2 + y2 = ab + bc
= b(a + c)
and = b x 2b [from(i)]
so= 2 b2
Hence x2, b2, y2 are in G.P.

#### Question 8.

If a, b, c are in G.P. and a, x, b, y, c are in A.P., prove that :
(i)$\frac { 1 }{ x } +\frac { 1 }{ y } =\frac { 2 }{ b }$
(ii)$\frac { a }{ x } +\frac { c }{ y } =2$

a, b, c are in G.P.
b2 = ac
a, x, b, y, c are in A.P.
2x = a + b and 2y = b + c

#### Question 9.

If a, b and c are in A.P. and also in G.P., show that: a = b = c.

a, b, c are in A.R
2 b = a + c ….(i)
Again, a, b, c are in G.P.

### Solutions of Concise Selina Publishers ICSE Maths Geometric Progression Exercise-11(D) for ICSE Maths Class 10th

#### Question 1.

Find the sum of G.P. :
(i) 1 + 3 + 9 + 27 +….to 12 terms.
(ii) 0.3 + 0.03 + 0.003 + 0.0003 +….to 8 terms.
(iii) $1-\frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } ....to\quad 9\quad terms$
(iv) $1-\frac { 1 }{ 3 } +\frac { 1 }{ { 3 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 3 } } ....to\quad n\quad terms$
(v) $\frac { x+y }{ x-y } +1+\frac { x-y }{ x+y } +....upto\quad n\quad terms$
(vi) $\sqrt { 3 } +\frac { 1 }{ \sqrt { 3 } } +\frac { 1 }{ 3\sqrt { 3 } } +....to\quad n\quad terms$

(i) 1 + 3 + 9 + 27 +….to 12 terms.
Here a = 1, r = 3 and n = 12

#### Question 2.

How many terms of the geometric progression 1 + 4 + 16 + 64 +…. must be added to get sum equal to 5461 ?
Solution:

Sn = 5461 and G.P. is
1 + 4 + 16 + 64 +…..
Here, a = 1, r = 4 (r > 1)

#### Question 3.

The first term of a G.P. is 27 and its 8th term is $\\ \frac { 1 }{ 81 }$. Find the sum of its first 10 terms.

First term of a G.P (a) = 27
T8 = $\\ \frac { 1 }{ 81 }$, n = 10
a = 27

#### Question 4.

A boy spends Rs 10 on first day, Rs 20 on second day, Rs 40 on third day and so on. Find how much, in all, will he spend in 12 days?

A boy spends Rs 10 on first day,
Rs 20 on second day
Rs 40 on third day and so on
G.P. is 10 + 20 + 40 +…. 12 terms
Here a = 10, r = 2 and n = 12 (r > 1)

#### Question 5.

The 4th and the 7th terms of a G.P. are $\\ \frac { 1 }{ 27 }$ and $\\ \frac { 1 }{ 729 }$ respectively. Find the sum of n terms of this G.P.

In a G.P.
T4 = $\\ \frac { 1 }{ 27 }$
=> $\\ \frac { 1 }{ 27 }$ = ar³

#### Question 6.

A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728 ; find its first term.

In a G.P.
Common ratio (r) = 3
Last term (l) = 486
Sum of its terms (Sn) = 728
Let a be the first term, then

#### Question 7

Find the sum of G.P. : 3, 6, 12, …… 1536.

G.P. is 3, 6, 12,….1536
Here a = 3, r = $\\ \frac { 6 }{ 3 }$ = 2

#### Question 8.

How many terms of the series 2 + 6 + 18 +…. must be taken to make the sum equal to 728 ?

G.P. is 2 + 6 + 18 +….
Here a = 2, r = $\\ \frac { 6 }{ 2 }$ = 3, Sn = 728

#### Question 9.

In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125 : 152. Find its common ratio.

In a G.P.
Sum of first 3 terms : Sum of 6 terms = 125 : 152

#### Question 10.

Find how many terms of G.P.$\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 }$… must be added to get the sum equal to $\\ \frac { 55 }{ 72 }$ ?

$\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 }$
Now,Sn = $\\ \frac { 55 }{ 72 }$

#### Question 11.

If the sum of 1 + 2 + 22 +…..+ 2n – 1 is 255, find the value of n.

1 + 2 + 22 +…..+ 2n – 1 = 255

#### Question 12.

Find the geometric mean between :
(i) $\\ \frac { 4 }{ 9 }$ and $\\ \frac { 9 }{ 4 }$
(ii) 14 and $\\ \frac { 7 }{ 32 }$
(iii) 2a and 8a3

(i) G.M between $\\ \frac { 4 }{ 9 }$ and $\\ \frac { 9 }{ 4 }$
$\sqrt { \frac { 4 }{ 9 } \times \frac { 9 }{ 4 } } =\sqrt { 1 } =1$

#### Question 13.

The sum of three numbers in G.P. is $\\ \frac { 39 }{ 10 }$ and their product is 1. Find the numbers.

Sum of three numbers in G.P. = $\\ \frac { 39 }{ 10 }$
and their product = 1
Let number be $\\ \frac { a }{ r }$, a, ar, then

#### Question 14.

The first term of a G.P. is – 3 and the square of the second term is equal to its 4th term. Find its 7th term.

In G.P.
T1 = – 3

#### Question 15

.Find the 5th term of the G.P. $\\ \frac { 5 }{ 2 }$, 1,….

Given G.P is $\\ \frac { 5 }{ 2 }$, 1,…..
Here a = $\\ \frac { 5 }{ 2 }$
and r = $\frac { 1 }{ \frac { 5 }{ 2 } }$
$\\ \frac { 2 }{ 2 }$

#### Question 16.

The first two terms of a G.P. are 125 and 25 respectively. Find

Given, First term = a = 125….(i)
and Second term = ar = 25…..(ii)
Now, Divide eq. (ii) by eq (i), we get

Question 17.

Find the sum of the sequence $- \frac { 1 }{ 3 }$, 1, – 3, 9,….upto 8 terms.

Here, First Term, a = $- \frac { 1 }{ 3 }$…(i)
and Second Term, ar = 1 …(ii)
Dividing eq. (i) by eq. (ii), we get

#### Question 18.

The first term of a G.P. in 27. If the 8th term be $\\ \frac { 1 }{ 81 }$, what will be the sum of 10 terms ?

Given, First term (a) = 27, n = 10

#### Question 19

.Find a G.P. for which the sum of first two terms is – 4 and the fifth term is 4 times the third term.

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

### ADDITIONAL QUESTION Geometric Progression Concise Solutions Chapter-11

#### Question 1.

Find the sum of n terms of the series :
(i) 4 + 44 + 444 + ….
(ii) 0.8 + 0.88 + 0.888 + ….

(i) 4 + 44 + 444 + ….
$=\frac { 4 }{ 9 } \left[ 9+99+999+.... \right]$

#### Question 2.

Find the sum of infinite terms of each of the following geometric progression:
(i)$1+\frac { 1 }{ 3 } +\frac { 1 }{ 9 } +\frac { 1 }{ 27 } +...$
(ii)$1-\frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } +...$
(iii)$\frac { 1 }{ 3 } +\frac { 1 }{ { 3 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 3 } } +...$
(iv)$\sqrt { 2 } -\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2\sqrt { 2 } } -\frac { 1 }{ 4\sqrt { 2 } } +...$
(v)$\sqrt { 3 } +\frac { 1 }{ \sqrt { 3 } } +\frac { 1 }{ 3\sqrt { 3 } } +\frac { 1 }{ 9\sqrt { 3 } } +...$

(i)$1+\frac { 1 }{ 3 } +\frac { 1 }{ 9 } +\frac { 1 }{ 27 } +...$ upto infinity
Sn = $\\ \frac { a }{ 1-r }$

#### Question 3.

The second term of a G.P. is 9 and sum of its infinite terms is 48. Find its first three terms.

In a G.P.
T2 = 9, sum of infinite terms = 48
Let a be the first term and r be the common ratio, therefore,
S = $\\ \frac { a }{ 1-r }$

#### Question 4.

Find three geometric means between $\\ \frac { 1 }{ 3 }$ and 432.

Let G1, G2 and G3 be three means between
$\\ \frac { 1 }{ 3 }$ and 432, then

#### Question 5.

Find :
(i) two geometric means between 2 and 16
(ii) four geometric means between 3 and 96.
(iii) five geometric means between $3 \frac { 5 }{ 9 }$ and $40 \frac { 1 }{ 2 }$.
Solution:

(i) Two G.M. between 2 and 16
Let G1 , and G1 be the G.M.,
then 2, G1, G2, 16

#### Question 7.

Find the numbers in G.P. whose sum is 52 and the sum of whose product in pairs is 624.

Sum of 3 numbers in G.P. = 52
and their product in pairs = 624
Let numbers be a, ar, ar²

#### Question 8.

The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.

Sum of three numbers in G.P. = 21
Sum of their squares = 189
Let three numbers be a, ar, ar², then
a + ar + ar² = 21
=> a( 1 + r + r²) = 21….(i)

Thanks