ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5

More Exercises

Question 1.
Find the sum of:
(i) 20 terms of the series 2 + 6 + 18 + …
(ii) 10 terms of series 1 + √3 + 3 + …
(iii) 6 terms of the GP. 1, \(– \frac { 2 }{ 3 } \) , \(\\ \frac { 4 }{ 9 } \), …
(iv) 20 terms of the GP. 0.15, 0.015, 0.0015,…
(v) 100 terms of the series 0.7 + 0.07 + 0.007 +…
(vi) 5 terms and n terms of the series \(1+\frac { 2 }{ 3 } +\frac { 4 }{ 9 } +…\)
(vii) n terms of the G.P. √7, √21, 3√7, …
(viii)n terms of the G.P. 1, – a, a², – a³, … (a ≠ – 1)
(ix) n terms of the G.P. x3, x5 , x7, … (x ≠ ±1).
Solution:
(i) 2 + 6 + 18 + … 20 terms
Here, a = 2, r = 3, n = 20, r > 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q1.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q1.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q1.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q1.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q1.6
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q1.7
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q1.8

Question 2.
Find the sum of the first 10 terms of the geometric series
√2 + √6 + √18 + ….
Solution:
√2 + √6 + √18 + ….
Here, a = √2 , r = √3, r > 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q2.1

Question 3.
Find the sum of the series 81 – 27 + 9….\(– \frac { 1 }{ 27 } \)
Solution:
Given
81 – 27 + 9….\(– \frac { 1 }{ 27 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q3.1

Question 4.
The nth term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then find its first term.
Solution:
In a G.P.
Tn =128
Sn = 255
r = 2,
Let a be the first term, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q4.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q4.2

Question 5.
If the sum of first six terms of any G.P. is equal to 9 times the sum of the first three terms, then find the common ratio of the G.P.
Solution:
Sum of first 6 terms of a G.P. = 9 x The of first 3 terms
Let a be the first term and r be the common ratio
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q5.1

Question 6.
A G.P. consists of an even number of terms. If the sum of all the terms is 3 times the sum of the odd terms, then find its common ratio.
Solution:
Let the G.P. be a, ar, ar2, … ar2n – 1
These are 2n in number, which is an even number
A.T.Q.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q6.1

Question 7.
(i) How many terms of the G.P. 3, 32, 33, … are needed to give the sum 120?
(ii) How many terms of the G.P. 1, 4, 16, … must be taken to have their sum equal to 341?
Solution:
In G.P.
(i) 3, 32, 33, …
Sum = 120, Here, a = 3, r = \(\frac { { 3 }^{ 2 } }{ 3 } \) = 3, r > 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q7.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q7.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q7.3

Question 8.
How many terms of the GP. 1, √2 > 2, 2 √2 , … are required to give a sum of 1023( √2 + 1)?
Solution:
GP. 1, √2 > 2, 2 √2 , …
Sum = 1023 (√2 + 1)
Here, a = 1, r = √2 . r > 1
Let number of terms be n, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q8.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q8.2

Question 9.
How many terms of the \(\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 } +…\) will make the sum \(\\ \frac { 55 }{ 72 } \) ?
Solution:
G.P. is \(\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 } +…\)
sum \(\\ \frac { 55 }{ 72 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q9.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q9.2

Question 10.
The 2nd and 5th terms of a geometric series are \(– \frac { 1 }{ 2 } \) and sum \(\\ \frac { 1 }{ 16 } \) respectively. Find the sum of the series upto 8 terms.
Solution:
In a G.P.
a2 = \(– \frac { 1 }{ 2 } \) and a5 = \(\\ \frac { 1 }{ 16 } \)
Let a be the first term and r be the common ratio
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q10.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q10.2

Question 11.
The first term of a G.P. is 27 and 8th term is \(\\ \frac { 1 }{ 81 } \) . Find the sum of its first 10 terms.
Solution:
In a G.P.
First term (a) = 27
a8 = 81
Let r be the common ratio, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q11.1

Question 12.
Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728
Solution:
Common ratio of a G.P. = 3
and last term = 486
and sum of terms = 728
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q12.1

Question 13.
In a G.P. the first term is 7, the last term is 448, and the sum is 889. Find the common ratio.
Solution:
In a GP.
First term (a) = 7, last term (l) = 448
and sum = 889
Let r be the common ratio, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q13.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q13.2

Question 14.
Find the third term of a G.P. whose common ratio is 3 and the sum of whose first seven terms is 2186.
Solution:
In a G.P.
Common ratio = 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q14.1

Question 15.
If the first term of a G.P. is 5 and the sum of first three terms is \(\\ \frac { 31 }{ 5 } \), find the common ratio.
Solution:
In a G.P.
First term (a) = 5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q15.2

Question 16.
The sum of first three terms of a GP. is to the sum of first six terms as 125 : 152. Find the common ratio of the GP.
Solution:
S3 ÷ S6 = 125 : 152
Let r be the common ratio and a be the first number, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q16.1

Question 17.
Find the sum of the products of the corresponding terms of the geometric progression 2, 4, 8, 16, 32 and 128, 32, 8, 2, \(\\ \frac { 1 }{ 2 } \)
Solution:
Sum of the product of corresponding terms of the G.M.s
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q17.1

Question 18.
Evaluate \(\sum _{ n=1 }^{ 50 }{ \left( { 2 }^{ n }-1 \right) } \)
Solution:
\(\sum _{ n=1 }^{ 50 }{ \left( { 2 }^{ n }-1 \right) } \)
Here n = 1, 2, 3,….,50
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q18.1

Question 19.
Find the sum of n terms of a series whose mth term is 2m + 2m.
Solution:
am = 2m + 2m
a1 = 21 + 2 x 1 = 2 + 2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q19.1

Question 20.
Sum the series
x(x + y) + x2 (x2 + y2) + x3 (x3 + y3) … to n terms.
Solution:
Given
Sn = x(x + y) + x2 (x2 + y2) + x3 (x3 + y3) … n terms
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q20.1

Question 21.
Find the sum of the series
1 + (1 + x) + (1 + x + x2) + … to n terms, x ≠ 1.
Solution:
1 + (1 + x) + (1 + x + x2) +… n terms, x ≠ 1
Multiply and divide by (1 – x)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q21.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q21.2

Question 22.
Find the sum of the following series to n terms:
(i) 7 + 77 + 777 + …
(ii) 8 + 88 + 888 + …
(iii) 0.5 + 0.55 + 0.555 + …
Solution:
(i) 7 + 77 + 777 + … n terms
= 7[1 + 11 + 111 + … n terms]
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q22.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q22.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5 Q22.3

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