If `S_1,S_2a n dS_3` be respectively the sum of n, 2n and 3n terms of a G.P., prove that `S_1(S_3-S_2)=(S_2-S_1)^2`

Explanation

`S_1​=(a(r^n−1))/(r-1)​`

`S_2​=(a(r^(2n)−1))/(r-1)​`

`S_3​=(a(r^(3n)−1))/(r-1)​`

`S_1​(S_3​−S_2​)=(a(r^n−1))/(r-1)​[(a(r^(3n)−1))/(r-1)​−(a(r^(2n)−1))/(r-1)​​]`

`=(a^2(r^n−1))/((r-1)​​^2)[r^(3n)−1−r^(2n)+1]`

`=(a^2(r^n−1))/((r-1)​​​^2)[r^(3n)−r^(2n)]`

`=(a^2)/(r-1)^2[​r^(2n)(r^n−1)^2]...(1)`

`(S_2​−S_1​)^2=[(a(r^(2n)−1)​)/(r-1)−(a(r^n−1))/(r-1)​]^2`

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