# If $a, b, c$ are in AP and $a^2$, $b^2$, $c^2$ in GP, and $a+b+c=\frac{3}{2}$, then value of $c$ is

$(a)\;\frac{1}{2\sqrt{2}}\qquad(b)\;\frac{1}{2}-\frac{1}{\sqrt{2}}\qquad(c)\;\frac{1}{2}+\frac{1}{\sqrt{2}}\qquad(d)\;\frac{1}{2}$

Answer : (c) $\;\frac{1}{2}+\frac{1}{\sqrt{2}}$
Explanation : $\;a+b+c=\frac{3}{2}$
$3b=\;\frac{3}{2}$
$b=\frac{1}{2}$
$a^2\;c^2=b^4=\frac{1}{16}\quad\;a+c=1$
$a^2\;(1-a)^2=\frac{1}{16}$
$a=\frac{1}{2}\pm\;\frac{1}{\sqrt{2}}\quad\;c=1-a=\frac{1}{2}\pm\frac{1}{\sqrt{2}}$
$a\;<\;b\;<\;c$
$c=\frac{1}{2}+\frac{1}{\sqrt{2}}.$