The sum of an infinite geometric series is 20, and the sum of the squares

Given: $\frac{\mathrm{a}}{1-\mathrm{r}}=20 \& \frac{\mathrm{a}^{2}}{1-\mathrm{r}^{2}}=100$

(Because on squaring both first term a and common ratio r will be squared.) To find: the series

$\mathrm{a}=20(1-\mathrm{r}) \ldots$ (i)

$\Rightarrow \frac{a^{2}}{1-r^{2}}=100=\frac{(20 \times(1-r))^{2}}{(1-r)(1+r)} \ldots($ from $(i))$

$\Rightarrow 100=400 \times \frac{1-r}{1+r}$

$\Rightarrow 100(1+r)=400(1-r)$

$\Rightarrow 100+100 r=400-400 r$

$\Rightarrow 100 r+400 r=400-100$

$\Rightarrow 500 r=300$

$\Rightarrow 5 r=3$

$\Rightarrow r=\frac{3}{5}$

Put this value of r in equation (i) we get

$a=20\left(1-\frac{3}{5}\right)=\frac{20 \times 2}{5}=8$

∴The infinite geometric series is:8, $\frac{24}{5}, \frac{72}{25}, \frac{216}{125}, \frac{648}{625}, \ldots \infty$