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

 Since, a, b, c  are  in AP.

$\therefore \mathrm{b}=\mathrm{a}+\mathrm{d},\mathrm{c}=\mathrm{a}+2\mathrm{d}\mathrm{, }$

where  d is a common difference,  d >0

Again, since  a²,b²,c²are  in GP.

$\therefore {\mathrm{a}}^2,(\mathrm{a}+\mathrm{d}{)}^2\text{ and }(\mathrm{a}+2\mathrm{d}{)}^2\text{ are in GP. }$

$\Rightarrow (\mathrm{a}+\mathrm{d}{)}^4={\mathrm{a}}^2(\mathrm{a}+2\mathrm{d}{)}^2$

or , $(\mathrm{a}+\mathrm{d}{)}^2=\pm \mathrm{a}(\mathrm{a}+2\mathrm{d})$

$\Rightarrow {\mathrm{a}}^2+{\mathrm{d}}^2+2\mathrm{ad}=\pm \left({\mathrm{a}}^2+2\mathrm{ad}\right)$

On  taking (+) sign,  d =0 (not  possible  as  o  < b < c)  
On  taking (-)  sign,   $2{\mathrm{a}}^2+4\mathrm{ad}+{\mathrm{d}}^2=0$

$\Rightarrow 2{\mathrm{a}}^2+4\mathrm{a}\left(\frac{1}{2}-\mathrm{a}\right)+{\left(\frac{1}{2}-\mathrm{a}\right)}^2=0$

   $\left[\because \mathrm{a}+\mathrm{b}+\mathrm{c}=\frac{3}{2}\Rightarrow \mathrm{a}+\mathrm{d}=\frac{1}{2}\right]$

  $\Rightarrow 4{\mathrm{a}}^2-4\mathrm{a}-1=0$

$\therefore \mathrm{a}=\frac{1}{2}\pm \frac{1}{\sqrt{2}}$

Here, $\mathrm{d}=\frac{1}{2}-\mathrm{a}>0$

So,  $\mathrm{a}<\frac{1}{2}$

Hence, $\mathrm{a}=\frac{1}{2}-\frac{1}{\sqrt{2}}$