Question

If S_n, S_2n, S_3n are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that S_n (S_3n – S_2n) = (S_2n – S_n)².

Answer 1

Let a and r be the 1^st term and common ratio of the G.P. respectively.

∴ S_n = `a((r^n - 1)/(r - 1)), S_(2n) = a((r^(2n) - 1)/(r - 1)), S_(3n) = a((r^(3n) - 1)/(r - 1))`

∴ S_2n – S_n = `a((r^(2n) - 1)/(r - 1)) - a((r^n - 1)/(r - 1))`

= `a/(r - 1)(r^(2n) - 1 - r^n + 1)`

= `a/(r - 1)(r^(2n) - r^n)`

= `ar^n/(r - 1) (r^n - 1)`

∴ S_2n – S_n = `r^n*(a(r^n - 1))/(r - 1)`     ....(i)

S_3n – S_2n = `a((r^(3n) - 1)/(r - 1)) - a((r^(2n) - 1)/(r - 1))`

= `a/(r - 1)(r^(3n) - 1 - r^(2n) + 1)`

= `a/(r - 1)(r^(3n) - r^(2n))`

= `a/(r - 1)*r^(2n)(r^n - 1)`

= `a*((r^n - 1)/(r - 1))*r^(2n)`

∴ S_n(S_3n – S_2n) = `[a*((r^n - 1)/(r - 1))][a*((r^n - 1)/(r - 1))r^(2n)]`

= `[r^n*(a(r^n - 1))/(r - 1)]^2`

∴ S_n(S_3n – S_2n) = (S_2n – S_n)²   ....[From (i)]