Suppose a, b, c are in A.P. and $a^2,b^2,c^2$ are in G.P. If a < b < c and a + b + c = 3/2, then the value of a is

a + c = 2b fi 3b = a + b + c = 3/2 fi b = ½

Thus, $a+c=1\text{ and }{a}^2{c}^2=b^4=1/16$

$\begin{array}{l}\Rightarrow a^2(1-a{)}^2=1/16\Rightarrow a(1-a)=\pm 1/4\\ \Rightarrow a^2-a\pm 1/4=0\\ \Rightarrow 4a^2-4a\pm 1=0\Rightarrow (2a-1{)}^2=0\text{ or }(2a-1{)}^2=2\\ \Rightarrow a=\frac{1}{2}\text{ or }a=\frac{1}{2}\pm \frac{1}{\sqrt{2}}\end{array}$

As a < b < c and $b=\frac{1}{2}$, we get a $=\frac{1}{2}-\frac{1}{\sqrt{2}}$.