Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive

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-

a_n = ar^n-1

As, a = 4 (given)

And a₅ – a₃ = \( \dfrac{32}{81}\) (given)

4r⁴ – 4r² = \( \dfrac{32}{81}\)

4r²(r² – 1) = \( \dfrac{32}{81}\)

r²(r² – 1) = \( \dfrac{8}{81}\)

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 = \( \dfrac{18}{162}= \dfrac{1}{9}\) or

y = \( \dfrac{8}{9}\)

So, r² = \( \dfrac{1}{9}\) or \( \dfrac{8}{9}\)

= \( \dfrac{1}{3}\)

We know that,

Sum of infinite, S_∞ = \(\dfrac{ a}{(1 – r)}\)

Where, a = 4, r = \(\dfrac{1}{3}\)

= \( \dfrac{12}{2}\)

= 6

Sum of infinite, S_∞ =\( \dfrac{ a}{(1 – r)}\)

Where, a = 4, r = \( \dfrac{2}{\sqrt{\dfrac{2}{3}}}\)

\( S_∞ = \dfrac{4}{\dfrac{2}{2\sqrt{3}}} \)

= \( \dfrac{12}{3-2\sqrt{2}}\)