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) ,

Additional Questions

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,…..

Answer 1

(i) 8, 24, 72, 216,……
Here, a = 8

Question 2

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

Answer 2

In G.P. 1, 4, 16, 64,….
Here first term (a) = 1
and common ratio (r) = $\\ \frac { 4 }{ 1 }$ = 4,
T₉ = ar^{n – 1} = 1 x 4^{9 – 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…….

Answer 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……

Answer 4

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, $1 \frac { 1 }{ 3 }$ ,……

Answer 5

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 …….

Answer 6

1, 2, 4, 8,……
Here, a = 1,r = $\\ \frac { 2 }{ 1 }$ = 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 6

Question 7

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

Answer 7

√5, 5, 5√5……
Here a = √5 and r = $\frac { 5 }{ \surd 5 }$ = √5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 7

Question 8

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

Answer 8

2², 2³, 2⁴,……
Here, a = 2², r = 2³ ÷ 2² = 2^{3 – 2} = 2¹ = 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 8

Question 9

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

Answer 9

√3 + 1, 1, $\frac { \surd 3-1 }{ 2 }$ ,…….
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 9

Question 10

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

Answer 10

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:
$\\ \frac { 2 }{ 27 }$ , $\\ \frac { 2 }{ 9 }$ , $\\ \frac { 2 }{ 3 }$ ,…..

Answer 11

G.P. is $\\ \frac { 2 }{ 27 }$ , $\\ \frac { 2 }{ 9 }$ , $\\ \frac { 2 }{ 3 }$ ,…..
a = $\\ \frac { 2 }{ 27 }$
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 11

Question 12

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

Answer 12

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, $\frac { 5 }{ \surd 3 } ,-\frac { 5 }{ 6 } ,....-\frac { 5 }{ 72 } ?$

Answer 1

– 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.

Answer 2

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.

Answer 3

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

Answer4

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.

Answer 5

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.

Answer 6

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.

Answer7

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.

Answer8

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.

Answer9

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.

Answer10

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