Explanation

(d)
Let the GP series be $a,ar,\ ar^{2},\ldots$
According to the given condition
$a + ar + ar^{2} + ... = a^{2} + a^{2}r^{2} + a^{2}r^{4} + ... = 3$
$\Rightarrow \ \ \frac{a}{1 - r} = \frac{a^{2}}{1 - r^{2}} = 3$
$\Rightarrow \ \ a = 3(1 - r)$ …(i)
and $a^{2} = 3(1 - r^{2})$ …(ii)
Eliminating $a$ from Eqs. (i) and (ii), we get
$\left\lbrack 3(1 - r) \right\rbrack^{2} = 3(1 - r^{2})$
$\Rightarrow \ 3(1 - r) = (1 + r)\ \ \ \ \ \ \ \ \ (\because r \neq 1)$
$\Rightarrow 4r = 2\ \Rightarrow r = \frac{1}{2}$
$\therefore\ a = 3(1 - r) = 3\left( 1 - \frac{1}{2} \right) = \frac{3}{2}$