If a, b, c are in A.P. and a², b², c² are in H.P., then which is of the following is/are possible?

Given that a, b, c are in A.P.

$\Rightarrow 2\mathrm{b}=\mathrm{a}+\mathrm{c}$                 ….. (1)

and ${\mathrm{a}}^2,{\mathrm{b}}^2,{\mathrm{c}}^2$ are in H.P.

$\begin{array}{l}\Rightarrow \frac{1}{{\mathrm{b}}^2}-\frac{1}{{\mathrm{a}}^2}=\frac{1}{{\mathrm{c}}^2}-\frac{1}{{\mathrm{b}}^2}\\ \Rightarrow \frac{(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})}{{\mathrm{b}}^2{\mathrm{a}}^2}=\frac{(\mathrm{b}-\mathrm{c})(\mathrm{b}+\mathrm{c})}{{\mathrm{b}}^2{\mathrm{c}}^2}\\ \Rightarrow {\mathrm{ac}}^2+{\mathrm{bc}}^2={\mathrm{a}}^2\mathrm{b}+{\mathrm{a}}^2\mathrm{c} [\because \mathrm{a}-\mathrm{b}=\mathrm{b}-\mathrm{c}]\\ \Rightarrow \mathrm{ac}(\mathrm{c}-\mathrm{a})+\mathrm{b}(\mathrm{c}-\mathrm{a})(\mathrm{c}+\mathrm{a})=0\\ \Rightarrow (\mathrm{c}-\mathrm{a})(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})=0\\ \Rightarrow \mathrm{c}-\mathrm{a}=0\text{ or }\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=0\end{array}$

For c = a, from (1), a = b = c

For (a + c)b + ca = 0, from (1),

$\begin{array}{c}2{\mathrm{b}}^2+\mathrm{ca}=0\\ \Rightarrow {\mathrm{b}}^2=\mathrm{a}\left(\frac{-\mathrm{c}}{2}\right)\end{array}$

This implies that a, b, -c/2 are in G.P..