If $a^2$, $b^2$ and $c^2$ are in $A.P$, prove that $\cot A$, $\cot B$ and $\cot C$ are also in $A.P$.
Given: $\;$ $a^2, \; b^2, \; c^2$ are in $A.P$
$\implies$ $2 b^2 = a^2 + c^2$
i.e. $\;$ $b^2 = a^2 +c^2 - b^2$ $\;\;\; \cdots \; (1)$
In any $\triangle ABC$,
$\cot A = \dfrac{b^2 + c^2 - a^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$, $\;$ $\cot B = \dfrac{c^2 + a^2 - b^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$, $\;$ $\cot C = \dfrac{a^2 + b^2 - c^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$
Now,
$\begin{aligned}
\cot A + \cot C & = \dfrac{b^2 + c^2 - a^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} + \dfrac{a^2 + b^2 - c^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} \\\\
& = \dfrac{2 b^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} \\\\
& = \dfrac{b^2}{2 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} \\\\
& = \dfrac{a^2 + c^2 - b^2}{2 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} \;\; \left[\text{in view of equation (1)}\right] \\\\
& = 2 \times \left(\dfrac{a^2 + c^2 - b^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}\right) \\\\
& = 2 \cot B
\end{aligned}$
i.e. $\;$ $\cot A + \cot C = 2 \cot B$
$\implies$ $\cot A, \; \cot B, \; \cot C$ are in $A.P$
Hence proved.