Explanation

(d)
Since, $a,\ b,\ c$ are in AP.
$\therefore b = a + d,c = a + 2d$,
Where $d$ is a common difference, $d > 0$
Again, since $a^{2},\ b^{2},\ c^{2}$ are in GP.
$\therefore\ a^{2},(a + d)^{2},\ (a + 2d)^{2}$ are in GP
$\Rightarrow (a + d)^{4} = a^{2}(a + 2d)^{2}$
or $(a + d)^{2} = \pm a(a + 2d)$
$\Rightarrow \ a^{2} + d^{2} + 2ad = \pm (a^{2} + 2ad)$
Taking (+) sign, $d = 0$ (not possible as $a < b < c)$
Taking $( - )$sign
$2a^{2} + 4ad + d^{2} = 0$
$\Rightarrow \ 2a^{2} + 4a\left( \frac{1}{2} - a \right) + \left( \frac{1}{2} - a \right)^{2} = 0$ $\left( \therefore a + b + c = \frac{3}{2}\ \Rightarrow a + d = \frac{1}{2} \right)$
$\Rightarrow \ 4a^{2} - 4a - 1 = 0$
$\therefore\ a = \frac{1}{2} \pm \frac{1}{\sqrt{2}}$
Here, $d = \frac{1}{2} - a > 0$
So, $a < \frac{1}{2}$
Hence, $a = \frac{1}{2} - \frac{1}{\sqrt{2}}$