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

It is given that a, b, c are in A.P. 

$\begin{array}{ll}\Rightarrow & 2\mathrm{b}=\mathrm{a}+\mathrm{c}\\ \Rightarrow & 2\mathrm{b}=\frac{3}{2}-\mathrm{b}[\because \mathrm{a}+\mathrm{b}+\mathrm{c}=\frac{3}{2}(\text{Given}\left)\right]\\ \Rightarrow & \mathrm{b}=\frac{1}{2}\Rightarrow \mathrm{a}+\mathrm{c}=1[\because \mathrm{a}+\mathrm{c}=2\mathrm{b}]\end{array}$

It is also given that  ${\mathrm{a}}^2,{\mathrm{b}}^2,{\mathrm{c}}^2$ are in G.P.

$\Rightarrow {\mathrm{b}}^2=\sqrt{{\mathrm{a}}^2{\mathrm{c}}^2}\Rightarrow {\mathrm{b}}^2=\pm \mathrm{ac}\Rightarrow \mathrm{ac}=\pm \frac{1}{4}[\because \mathrm{b}=\frac{1}{2}]$

CASE I: When ac =½  In this case, we have 

               $\begin{array}{rl} & \mathrm{ac}=\frac{1}{4}\text{and }\mathrm{a}+\mathrm{c}=1\\ \therefore & \mathrm{a}+\frac{1}{4\mathrm{a}}=1\Rightarrow (2\mathrm{a}-1{)}^2=0\Rightarrow \mathrm{a}=\frac{1}{2}\end{array}$

               Thus, we have a = c =b =½ which is not possible as a  <b<c is given

CASE II: When ac = -½  In this case, we have

               $\begin{array}{rl} & \mathrm{ac}=-\frac{1}{4}\text{and }\mathrm{a}+\mathrm{c}=1\\ & \Rightarrow \mathrm{a}-\frac{1}{4\mathrm{a}}=1\\ & \Rightarrow 4{\mathrm{a}}^2-4\mathrm{a}-1=0\\ & \Rightarrow \mathrm{a}=\frac{4\pm \sqrt{16+16}}{8}=\frac{1}{2}\pm \frac{1}{\sqrt{2}}\\ & \text{ If }\mathrm{a}=\frac{1}{2}+\frac{1}{\sqrt{2}}\text{' then}\mathrm{a}+\mathrm{c}=1\Rightarrow \mathrm{c}=\frac{1}{2}-\frac{1}{\sqrt{2}}\end{array}$

                But, this is not possible as a <c Thus, we have  $\mathrm{a}=\frac{1}{2}-\frac{1}{\sqrt{2}}$