$\because a, b, c$ are in A.P. then
$$
a+c=2 b
$$
also it is given that,
$$
\begin{aligned}
&a+b+c=\frac{3}{4} \\
&\Rightarrow 2 b+b=\frac{3}{4} \Rightarrow b=\frac{1}{4}
\end{aligned}
$$
Again it is given that, $a^2, b^2, c^2$ are in G.P. then
$$
\left(b^2\right)^2=a^2 c^2 \Rightarrow a c=\pm \frac{1}{16}
$$
From (1), (2) and (3), we get;
$$
a \pm \frac{1}{16 a}=\frac{1}{2} \Rightarrow 16 a^2-8 a \pm 1=0
$$
Case I: $16 a^2-8 a+1=0$
$\Rightarrow a=\frac{1}{4}$ (not possible as $\left.a < b\right)$
Case II: $16 a^2-8 a-1=0 \Rightarrow a=\frac{8 \pm \sqrt{128}}{32}$
$$
\begin{aligned}
&\Rightarrow a=\frac{1}{4} \pm \frac{1}{2 \sqrt{2}} \\
&\therefore a=\frac{1}{4}-\frac{1}{2 \sqrt{2}} \quad(\because a < b)
\end{aligned}
$$