Question -
Answer -
Let ‘a’ be the firstterm of GP and ‘r’ be the common ratio.
We know that nth termof 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 withy
81y(y-1) = 8
81y2 –81y – 8 = 0
Using the formula ofthe 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∞ = 4/ (1 – (1/3))
= 4 / ((3-1)/3)
= 4 / (2/3)
= 12/2
= 6
Sum of infinite, S∞ =a/(1 – r)
Where, a = 4, r = 2√2/3
S∞ = 4/ (1 – (2√2/3))
= 12 / (3 – 2√2)