# If S denotes the sum of an infinite G.P. S1 denotes the sum of the squares of its terms,

Question:

If $S$ denotes the sum of an infinite G.P. $S_{1}$ denotes the sum of the squares of its terms, then prove that the first term and common ratio are respectively $\frac{2 S S_{1}}{S^{2}+S_{1}}$ and $\frac{S^{2}-S_{1}}{S^{2}+S_{1}}$.

Solution:

$\mathrm{S}=\frac{a}{(1-r)}$    ...(i)

And, $\mathrm{S}_{1}=\frac{a^{2}}{\left(1-r^{2}\right)}$

$\Rightarrow \mathrm{S}_{1}=\frac{a^{2}}{(1-r)(1+r)} \quad \cdots \cdots$ (ii)

Now, putting the value of $a$ in equation (ii) from equation (i):

$\mathrm{S}_{1}=\frac{\mathrm{S}^{2}(1-r)^{2}}{(1-r)(1+r)}$

$\Rightarrow \mathrm{S}_{1}=\frac{\mathrm{S}^{2}(1-r)}{(1+r)}$

$\Rightarrow \mathrm{S}_{1}(1+r)=\mathrm{S}^{2}(1-r)$

$\Rightarrow r\left(\mathrm{~S}_{1}+\mathrm{S}^{2}\right)=\mathrm{S}^{2}-\mathrm{S}_{1}$

$\Rightarrow r=\frac{\left(\mathrm{S}^{2}-\mathrm{S}_{1}\right)}{\left(\mathrm{S}_{1}+\mathrm{S}^{2}\right)}$

Putting the value of $r$ in equation $(\mathrm{i})$ :

$\Rightarrow a=\mathrm{S}(1-r)$

$\Rightarrow a=\mathrm{S}\left(1-\frac{\left(\mathrm{S}^{2}-\mathrm{S}_{1}\right)}{\left(\mathrm{S}_{1}+\mathrm{S}^{2}\right)}\right)$

$\Rightarrow a=\mathrm{S}\left(\frac{\left(\mathrm{S}_{1}+\mathrm{S}^{2}\right)-\left(\mathrm{S}^{2}-\mathrm{S}_{1}\right)}{\left(\mathrm{S}_{1}+\mathrm{S}^{2}\right)}\right)$

$\Rightarrow a=\frac{2 \mathrm{SS}_{1}}{\left(\mathrm{~S}_{1}+\mathrm{S}^{2}\right)}$