If the sum of an infinite decreasing G.P. is 3 and the sum of the squares of its term is

Let us take a G.P. whose first term is a and common difference is r.

$\therefore S_{\infty}=\frac{a}{1-r}$

$\Rightarrow \frac{a}{1-r}=3$    ...(i)

And, sum of the terms of the G.P. $a^{2},(a r)^{2},\left(a r^{2}\right)^{2}, \ldots \infty:$

$S_{\infty}^{\prime}=\frac{a^{2}}{1-r^{2}}$

$\Rightarrow \frac{a^{2}}{1-r^{2}}=\frac{9}{2}$     ....(ii)

$\Rightarrow 2 a^{2}=9\left(1-r^{2}\right)$      $[$ From (i) $]$

$\Rightarrow 2[3(1-r)]^{2}=9-9 r^{2}$

$\Rightarrow 18-9+18 r^{2}+9 r^{2}-36 r=0$

$\Rightarrow 27 r^{2}-36 r+9=0$

$\Rightarrow 3\left(9 r^{2}-12 r+3\right)=0$

$\Rightarrow 9 r^{2}-9 r-3 r+3=0$

$\Rightarrow 9 r(r-1)-3(r-1)=0$

$\Rightarrow(9 r-3)(r-1)=0$

$\Rightarrow r=\frac{1}{3}$ and $r=1$

But, $r=1$ is not possible.

$\therefore r=\frac{1}{3}$

Now, putting $r=\frac{1}{3}$ in $\frac{\mathrm{a}}{1-\mathrm{r}}=3:$

$a=3\left(1-\frac{1}{3}\right)$

$\Rightarrow a=3 \times \frac{2}{3}=2$