If $\; A, \; B, \; C \;$ in $\;$ $\triangle ABC$ $\;$ are in A.P. and the sides $\; a, \; b, \; c \;$ are in G.P., then show that $\; a^2, \; b^2, \; c^2 \;$ are in A.P.
$\because \;$ $A, \; B, \; C \;$ are in A.P. $\implies$ $2 B = A + C$ $\;\;\; \cdots \; (1)$
$\because \;$ $a, \; b, \; c \;$ are in G.P. $\implies$ $b^2 = a c$ $\;\;\; \cdots \; (2)$
In $\triangle ABC$, $\;$ $A + B + C = \pi$
i.e. $\;$ $A + C = \pi - B$ $\;\;\; \cdots \; (3)$
In view of equation $(1)$, equation $(3)$ becomes,
$2 B = \pi - B$ $\implies$ $3 B = \pi$ $\implies$ $B = \dfrac{\pi}{3}$ $\; \; \; \cdots \; (4)$
By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$
i.e. $\;$ $b^2 = c^2 + a^2 - 2 c a \cos B$
i.e. $\;$ $b^2 = c^2 + a^2 - 2 b^2 \cos \left(\dfrac{\pi}{3}\right)$ $\;\;\;$ [by equations $(2)$ and $(4)$]
i.e. $\;$ $b^2 = c^2 + a^2 - 2 b^2 \times \dfrac{1}{2}$
i.e. $\;$ $b^2 = c^2 + a^2 - b^2$
i.e. $\;$ $2 b^2 = a^2 + c^2$
$\implies$ $a^2, \; b^2, \; c^2 \;$ are in A.P.