Question

If $S_1,S_2,S_3$ 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

From given we can write:

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

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

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

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

$=\frac{a^2(r^n-1)}{(r-1{)}^2}[r^{3n}-1-r^{2n}+1]$

$=\frac{a^2(r^n-1)}{(r-1{)}^2}[r^{3n}-r^{2n}]$

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

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

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

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

From (1) and (2), we have:

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