Race against time!

since $a , b , c$ are in an A.P,

$a + c = 2b$

So, $a + b + c = 3b = \frac{3}{2}$

$b = \frac{1}{2}$

let $-d$ be the common difference, where $d > 0$ .

so , $a = \frac{1}{2} + d$ and $c = \frac{1}{2} - d$ .

since $a^2 , b^2 , c^2$ are in G.P ,

$b^4 = (a^2)(c^2)$

so , $\frac{1}{16} = (\frac{1}{4} - d^2)^2$

$\frac{1}{4} - d^2 = +\frac{1}{4}$ OR $-\frac{1}{4}$

hence , $d^2 = 0$ OR $\frac{1}{2}$ .

$d = \frac{1}{\sqrt{2}}$ and hence $\boxed{ a = \frac{1}{2} + \frac{1}{\sqrt{2}} }$

a simpler solution given a+b+c=3÷2 as a b c are in ap 2b=a+c substituting in 1st equation we have b=1÷2 aleardy given in question a>b only option more than 0.5 is 1÷2 +1÷root 2 so that is the answer way to solve in 10 seconds