Indian Institute of Foreign Trade 2008 Solved Paper

Let us assume that the common difference of the A.P. is 'd'.
Then, we can say that $\mathit{a}=\mathit{b}-\mathit{d},\mathit{c}=\mathit{b}+\mathit{d}$
It is given that $\mathit{a}+\mathit{b}+\mathit{c}=\frac{3}{2}$. i.e. b = 1/2.
It is given that ${\mathit{a}}^2,{\mathit{b}}^2$ and ${\mathit{c}}^2$ are in Geometric Progression. Hence, we can say that
${\mathit{b}}^4={\mathit{a}}^2\ast {\mathit{c}}^2$
${\mathit{b}}^4={\left({\mathit{b}}^2-{\mathit{d}}^2\right)}^2$
$\Rightarrow \left({\mathit{b}}^2+{\mathit{d}}^2-{\mathit{b}}^2\right)\left({\mathit{b}}^2-{\mathit{d}}^2+{\mathit{b}}^2\right)=0$
Therefore,$\left({\mathit{b}}^2-{\mathit{d}}^2+{\mathit{b}}^2\right)=0$ i.e. $\mathit{d}=\frac{1}{\sqrt{2}}$
Hence, $\mathit{a}=\mathit{b}-\mathit{d}=\frac{1}{2}-\frac{1}{\sqrt{2}}$