Geometric Progression Concise Solutions Chapter-11 ICSE Maths

Selina Solutions Chapter-11 Geometric Progression

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 1.1

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 = 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…….

Answer 3

G.P. is 1 , √3, 3, 3√3
Here first term (a) = 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 3.1

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…….
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 4

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 } ,…..
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 5.1

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 = 23 – 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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 11.1

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)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B 1
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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B 2

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.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B 3

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B 5

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B 6

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B 7

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B 8

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B 9

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B 10

 

Geometric Progression Exercise-11(C)Concise Solutions 

 

Question 1.

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

Answer 1

√2, 2, 2√2,……32
Here a = √2
r = \frac { 2 }{ \surd 2 } =\surd 2
and l =32
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C 1

Question 2.

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

Answer 2

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C 2

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.

Answer 3

\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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C 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

Answer 4

In a G.P
Tp = a,
and Tq = b,
so Tr = c
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C 4

Question 5.

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

Answer 5 

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.

Answer 6

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.

Answer 7

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

Answer 8

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C 8
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C 8.1

Question 9.

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

Answer 9

a, b, c are in A.R
2 b = a + c ….(i)
Again, a, b, c are in G.P.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C 9

 

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

Answer 1

(i) 1 + 3 + 9 + 27 +….to 12 terms.
Here a = 1, r = 3 and n = 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 1.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 1.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 1.5

Question 2.

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

Answer 2

Sn = 5461 and G.P. is
1 + 4 + 16 + 64 +…..
Here, a = 1, r = 4 (r > 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 2

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.

Answer 3

First term of a G.P (a) = 27
T8 = \\ \frac { 1 }{ 81 } , n = 10
a = 27
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 3

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?

Answer 4

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)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 4

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.

Answer 5

In a G.P.
T4 = \\ \frac { 1 }{ 27 }
=> \\ \frac { 1 }{ 27 }  = ar³
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 5

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.

Answer 6

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
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 6

Question 7

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

Answer 7

G.P. is 3, 6, 12,….1536
Here a = 3, r = \\ \frac { 6 }{ 3 }  = 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 7.1

Question 8.

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

Answer 8

G.P. is 2 + 6 + 18 +….
Here a = 2, r = \\ \frac { 6 }{ 2 }  = 3, Sn = 728
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 8

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.

Answer 9

In a G.P.
Sum of first 3 terms : Sum of 6 terms = 125 : 152
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 9

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 }  ?

Answer 10

\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 }
Let n terms be added
Now,Sn = \\ \frac { 55 }{ 72 }
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 10.2

Question 11.

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

Answer 11

1 + 2 + 22 +…..+ 2n – 1 = 255
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 11

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

Answer 12

(i) G.M between \\ \frac { 4 }{ 9 }  and \\ \frac { 9 }{ 4 }
\sqrt { \frac { 4 }{ 9 } \times \frac { 9 }{ 4 } } =\sqrt { 1 } =1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 12.1

Question 13.

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

Answer 13

Sum of three numbers in G.P. = \\ \frac { 39 }{ 10 }
and their product = 1
Let number be \\ \frac { a }{ r } , a, ar, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 13

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.

Answer 14

In G.P.
T1 = – 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 14

Question 15

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

Answer 15

Given G.P is \\ \frac { 5 }{ 2 } , 1,…..
Here a = \\ \frac { 5 }{ 2 }
and r = \frac { 1 }{ \frac { 5 }{ 2 } }
\\ \frac { 2 }{ 2 }
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 15

Question 16.

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

Answer 16

Given, First term = a = 125….(i)
and Second term = ar = 25…..(ii)
Now, Divide eq. (ii) by eq (i), we get
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 16.1

Question 17.

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

Answer 17

Here, First Term, a = - \frac { 1 }{ 3 } …(i)
and Second Term, ar = 1 …(ii)
Dividing eq. (i) by eq. (ii), we get
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 17

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 ?

Answer 18

Given, First term (a) = 27, n = 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 18
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 18.1

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.

Answer 19

Let a be the first term and r be the common ratio of the G.P.
According to the given conditions,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 19

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 + ….

Answer 1

(i) 4 + 44 + 444 + ….
=\frac { 4 }{ 9 } \left[ 9+99+999+.... \right]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 1.1

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 } } +...

Answer 2

(i)1+\frac { 1 }{ 3 } +\frac { 1 }{ 9 } +\frac { 1 }{ 27 } +... upto infinity
Sn = \\ \frac { a }{ 1-r }
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 2.1

Question 3.

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

Answer 3

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 }
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 3

Question 4.

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

Answer 4

Let G1, G2 and G3 be three means between
\\ \frac { 1 }{ 3 }  and 432, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 4.1

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:

Answer 5

(i) Two G.M. between 2 and 16
Let G1 , and G1 be the G.M.,
then 2, G1, G2, 16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 5.2

 

Question 7.

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

Answer 7

Sum of 3 numbers in G.P. = 52
and their product in pairs = 624
Let numbers be a, ar, ar²
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 7.1

Question 8.

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

Answer 8

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)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 8
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions 8.1

End of Geometric Progression Concise Solutions Chapter-11

Chapter Wise Concise Maths Solution for ICSE Class 10th

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  10. Arithmetic Progression
  11. Geometric Progression ( Currently Open )
  12. Reflection
  13. Section And Mid Point Formula 
  14. Equation of a Line
  15. Similarity (With Applications to Maps and Models )
  16. Loci (Locus and its Constructions)
  17. Circles
  18. Tangents and Intersecting Chords
  19. Constructions ( Circles)
  20. Cylinder, Cone and Spheres (Surface Area and Volume )
  21. Trigonometrical Identities
  22. Heights and Distances
  23. Graphical Representation (Histograms and Ogives)
  24. Measures of Central Tendency (Mean , Median, Quartiles and Mode)
  25. Probability

 

Other Subject Chapter-Wise Solution for ICSE

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