Question

If $S_1,S_2$ and $S_3$ are 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 of G.P. and $r$ be the common ratio.

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

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

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

$S_1(S_3-S_2)=a^2\frac{(r^{4n}-2r^{3n}+r^{2n})}{{(r-1)}^2}$

${(S_2-S_1)}^2=a^2\frac{{(r^{2n}-r^n)}^2}{{(r-1)}^2}=a^2\frac{(r^{4n}-2r^{3n}+r^{2n})}{{(r-1)}^2}$

Thus $\text{L.H.S.}=\text{R.H.S.}$

Hence, proved.