Find the sum of t

Let a denote 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 = 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 have-

y =

⇒

∴ y = 18/162 = 1/9 or y = 144/162 = 8/9

⇒ r² = 1/9 or 8/9

∴

As GP is decreasing and all the terms are positive so we will consider only those values of r which are positive and |r|<1

∴ r =

∵ Sum of infinite GP = ,where a is the first term and k is the common ratio.

Note: We can only use the above formula if |k|<1

∴ the sum of respective GPs are –

S₁ = {sum of GP for r = 1/3}

S₂ = {sum of GP for r = (2√2)/3}