Question

If $S_1,S_2$ and $S_3$ and the sum of first $n,2n$ and $3n$ terms of a geometric series respectively, then prove that $S_1(S_3-S_2)={(S_2-S_1)}^2$

Answer 1

Let '$a$' be the first term.
'$r$' be the common ratio.
Given, $S_1,S_2,S_3$ are sum of first $n,2n$ and $3n$
$S_1=\frac{a(r^n-1)}{r-1}$
$S_2=\frac{a(r^{2n}-1)}{r-1}$
$S_3=\frac{a(r^{3n}-1)}{r-1}$
$S_3-S_2=\frac{a(r^{3n}-1)}{r-1}-\frac{a(r^{2n}-1)}{r-1}$
$=\frac{a}{r-1}(r^{3n}-1-(r^{2n}-1))$
$=\frac{a}{r-1}(r^{3n}-1-r^{2n}+1)=\frac{a}{r-1}(r^{3n}-r^{2n})$
$=\frac{a}{r-1}{r}^{2n}(r^n-1)$
$S_1(S_2-S_3)=\frac{a(r^n-1)}{r-1}\frac{a}{r-1}{r}^{2n}(r^n-1)$
$=\frac{a^2{r}^{2n{(r^n-1)}^2}}{{(r-1)}^2}$ ..........(1)
$S_2-S_1=a\frac{(r^{2n}-1)}{r-1}-\frac{a(r^n-1)}{r-1}$
$=\frac{a}{r-1}(r^{2n}-1-r^n+1)=\frac{a}{r-1}(r^{2n}-r^n)$
$=\frac{a}{r-1}{r}^n(r^n-1)$
$={(S_2-S_1)}^2=\frac{a^2}{{(r-1)}^2}{r}^{2n}{(r^n-1)}^2$ ..........(2)
From (1) and (2), we have
$S_1(S_3-S_2)={(S_2-S_1)}^2$