Let $r$ be the radius of the circle. We know that $r=\frac{1}{2}$ a cot $\left(\frac{\pi}{n}\right)$ for a regular polygon of $n$ sides of length a circumscribed about this circle.
Therefore, $a=2 r \tan \left(\frac{\pi}{n}\right)$. Now if $a_1, a_2$ and $a_3$ are the sides of the equilateral triangle, square and regular hexagon respectively, then
$a_1=2 r \tan \frac{\pi}{3}=2 \sqrt{3} r, a_2=2 r \tan \frac{\pi}{4}=2 r$ and $a_3=2 r \tan \frac{\pi}{6}=\frac{2}{\sqrt{3}} r$
we see that $a_1 a_3=(2 r)^2=a_2{ }^2$, from which it follows that $a_1, a_2$ and $a_3$ are in G.P.