Question

If $S_1,S_2,S_3$ be respectively the sums of $n,2n,3n$ terms of a $G.P.$, then prove that
$S_1(S_3-S_2)={(S_2-S_1)}^2$

Answer 1

$S_1=\frac{a(r^n-1)}{r-1}$, $S_2=\frac{a(r^{2n}-1)}{r-1}$, $S_2=\frac{a(r^{3n}-1)}{r-1}$
$\therefore S_3-S_2=\frac{a}{r-1}(r^{3n}-r^{2n})=\frac{a(r^n-1)}{r-1}.r^{2n}$
$\therefore S_1(S_3-S_2)=\frac{a(r^n-1)}{r-1}.\frac{a(r^n-1)}{r-1}{r}^{2n}$
$={\left[\frac{a(r^n-1)}{r-1}{r}^n\right]}^2$
$S_2-S_1=\frac{a}{r-1}(r^{2n}-r^n)=\frac{a(r^n-1)}{r-1}{r}^n$
$\therefore S_1(S_3-S_2)={(S_2-S_1)}^2$, by $\left(1\right)$ and $\left(2\right)$.