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

Let  $\text{a }=\text{ b }–\text{ d ad c }=\text{ b }+\text{ d},\text{ then a }+\text{ b }+\text{ c}=\frac{3}{2}\Rightarrow b=\frac{1}{2}$
  The numbers are  $\frac{\text{1}}{\text{2}}\text{-d,}\frac{\text{1}}{\text{2}}\text{,}\frac{\text{1}}{\text{2}}\text{+d}\left[\text{d > 0\hspace{0.17em}as \hspace{0.17em}a < b< c}\right]$
Now,  ${\text{a}}^{\text{2}}{\text{,b}}^{\text{2}}{\text{,c}}^{\text{2}}$ are in G.P.  $\Rightarrow {\left(b^2\right)}^2-a^2{c}^2\Rightarrow {\left(\frac{1}{2}\right)}^2={(\frac{1}{2}-d)}^2{(\frac{1}{2}+d)}^2$
 
$\Rightarrow \frac{1}{16}={(\frac{1}{4}-d^2)}^2\Rightarrow \frac{1}{4}-d^2=\underset{\_}{+}\frac{1}{4}\Rightarrow d=\underset{\_}{+}\frac{1}{\sqrt{2}}\Rightarrow d=\frac{1}{\sqrt{2}}(\because d>0)$