Question

if $S_n$ denotes the sum to $n$ terms of a $G.P$ show that $(S_{10}-S_{20}{)}^2=S_{10}(S_{30}-S_{20})$

Answer 1

$S_n=$ Sum of n terms in a G.P.

$S_n=\frac{a(r^n-1)}{r-1}(S_{10}-S_{20}{)}^2={(\frac{a(r^{10}-1)}{r-1}-\frac{a(r^{20}-1)}{r-1})}^2=\left(\frac{a(r^{10}-r^{20}{)}^2}{r-1}\right)=\frac{a^2(r^{20}-2r^{30}+r^{40})}{(r-1{)}^2}=\frac{a^2\left(r^{20}\right)(r^{20}-2r^{10}+1)}{(r-1{)}^2}=\frac{a^2{r}^{20}(r^{10}-1{)}^2}{(r-1{)}^2}=\frac{a^2{r}^{20}(r^{10}-1)(r^{10}-1)}{(r-1{)}^2}=\frac{a^2(r^{30}-r^{10})(r^{10}-1)}{(r-1{)}^2}=\frac{a(r^{10}-1)}{(r-1)}\left[\frac{a^2(r^{30}-r^{10})}{r-1}\right]=S_{10}\left[\frac{a(r^{30}-1-r^{20}-1)}{r-1}\right](S_{10}-S_{20}{)}^2=S_{10}[S_{30}-S_{20}]$