If $a,b,c,$ are in A.P and  $a^2,b^2,c^2$ are in H.P, then

We have  $2b=a=c$

$\begin{array}{l}\text{and }{b}^2=\frac{2a^2{c}^2}{a^2+c^2}\\ {\left(\frac{a+c}{2}\right)}^2=\frac{2a^2{c}^2}{a^2+c^2}\\ \Rightarrow {(a+c)}^2(a^2+c^2)=8a^2{c}^2\\ \Rightarrow {(a^2+c^2)}^2+2ac(a^2+c^2)-8a^2{c}^2=0\\ \Rightarrow (a^2+c^2+4ac)(a^2+c^2-2ac)=0\\ [{(a+c)}^2+2ac]{[a-c]}^2=0\\ a-c=0\Rightarrow a=c\\ \text{Then }a=b=c\end{array}$