If $a$ and $x$ are positive integers such that $x<a$ and $\sqrt{a-x}, \sqrt{x}, \sqrt{a+x}$ are in A.P., then least possible value of $a$ is

We have $2\sqrt{x}=\sqrt{a+x}+\sqrt{a-x}$
$\begin{array}{l}\Rightarrow 2\sqrt{x}=\frac{2x}{\sqrt{a+x}-\sqrt{a-x}}\Rightarrow \sqrt{a+x}-\sqrt{a-x}=\sqrt{x}\\ a+x+a-x-2\sqrt{a^2-x^2}=x\Rightarrow 2a-x=2\sqrt{a^2-x^2}\\ \Rightarrow 4a^2+x^2-4ax=4a^2-4x^2\Rightarrow 5x^2=4ax\\ \therefore a=\frac{5}{4}x\end{array}$
As $a$ and $x$ are positive integers, least possible value of $x$ is $4$ and that of $a$ is $5.$