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
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) $