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

Since $S_n$ is the sum of the $n$ terms of a G.P., then,

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

Simplifying the LHS of ${\left(S_{10}-S_{20}\right)}^2=S_{10}\left(S_{30}-S_{20}\right)$,

${\left(S_{10}-S_{20}\right)}^2={\left(\frac{a\left(1-r^{10}\right)}{1-r}-\frac{a\left(1-r^{20}\right)}{1-r}\right)}^2$

$={\left(\frac{a\left(1-r^{10}\right)-a\left(1-r^{20}\right)}{1-r}\right)}^2$

$={\left(\frac{a-a{r}^{10}-a+a{r}^{20}}{1-r}\right)}^2$

$={\left(\frac{a{r}^{20}-a{r}^{10}}{1-r}\right)}^2$

$=a^2{\left(r^{10}\left(\frac{r^{10}-1}{1-r}\right)\right)}^2$

$=a^2{r}^{20}{\left(\frac{r^{10}-1}{1-r}\right)}^2$

$=a^2{r}^{20}{\left(\frac{1-r^{10}}{1-r}\right)}^2$

Simplifying the RHS of ${\left(S_{10}-S_{20}\right)}^2=S_{10}\left(S_{30}-S_{20}\right)$,

$S_{10}\left(S_{30}-S_{20}\right)=\frac{a\left(1-r^{10}\right)}{1-r}\left(\frac{a\left(1-r^{30}\right)}{1-r}-\frac{a\left(1-r^{20}\right)}{1-r}\right)$

$=\frac{a\left(1-r^{10}\right)}{1-r}\left(\frac{a-a{r}^{30}-a+a{r}^{20}}{1-r}\right)$

$=\frac{a\left(1-r^{10}\right)}{1-r}\left(\frac{a{r}^{20}\left(1-r^{10}\right)}{1-r}\right)$

$=a^2{r}^{20}{\left(\frac{1-r^{10}}{1-r}\right)}^2$

This shows that $LHS=RHS$.

Hence proved.