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Home » Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is equal to 32/81.
Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is equal to 32/81.
CBSE | Class 11 | Exercise 20.4 | Geometric Progressions | Maths | RD Sharma
Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is equal to 32/81.

 Solution:

Let ‘a’ be the first term of GP and ‘r’ be the common ratio.

We know that nth term of a GP is given by-

an = arn-1

As, a = 4 (given)

And a5 – a3 = 32/81 (given)

4r4 – 4r2 = 32/81

4r2(r2 – 1) = 32/81

r2(r2 – 1) = 8/81

Let us denote r2 with y

81y(y-1) = 8

81y2 – 81y – 8 = 0

Using the formula of the quadratic equation to solve the equation, we get

RD Sharma Solutions for Class 11 Maths Chapter 20 – Geometric Progressions image - 12

y = 18/162 = 1/9 or

y = 144/162

= 8/9

So, r2 = 1/9 or 8/9

= 1/3 or 2√2/3

We know that,

Sum of infinite, S∞ = a/(1 – r)

Where, a = 4, r = 1/3

{{S}_{\infty }}~=\text{ }4\text{ }/\text{ }\left( 1-\left( 1/3 \right) \right)

=\text{ }4\text{ }/\text{ }\left( \left( 3-1 \right)/3 \right)

=\text{ }4\text{ }/\text{ }\left( 2/3 \right)

=\text{ }12/2

=\text{ }6

Sum of infinite, S∞ = a/(1 – r)

Where, a = 4, r = 2√2/3

{{S}_{\infty }}~=\text{ }4\text{ }/\text{ }\left( 1-\left( 2\surd 2/3 \right) \right)

=\text{ }12\text{ }/\text{ }\left( 3-2\surd 2 \right)

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