Question

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

Answer 1

let series is

$a,ar,.....a{r}^{n-1},a{r}^n,......a{r}^{2n-1},a{r}^{3n},......a{r}^{3n-1}$

Sum of $n$ terms

$S_1=\frac{a(1-r^n)}{1-r}$

Sum of $2n$ terms

$S_2=\frac{a(1-r^{2n})}{1-r}$

Sum of $3n$ terms

$S_3=\frac{a(1-r^{3n})}{1-r}$

New LHS

$=S_1(S_3-S_2)$

$\frac{a(1-r^n)}{(1-r)}\left(\frac{a}{1-r}\right(1-r^{3n}-1+r^{2n}\left)\right)$

$=\frac{a^2}{{(1-r)}^2}{r}^{2n}{(1-r^n)}^2.....\left(1\right)$

RHS

${(S_2-S_1)}^2={\left[\frac{a}{1-r}(1-r^{2n}-1+r^n)\right]}^2$

$=\frac{a^2}{{(1-r)}^2}{\left[(r^n-r^{2n})\right]}^2=\frac{a^2}{{(1-r)}^2}{(1-r^n)}^2.....\left(2\right)$

LHS=RHS

$S_1(S_3-S_2)={(S_2-S_1)}^2$