$\text{If }\frac{1}{\beta +\gamma },\frac{1}{\alpha +\gamma },\frac{1}{\alpha +\beta }\text{ are in A}\text{.P, Prove that }{{\alpha }^{2}},{{\beta }^{2}},{{\gamma }^{2}}\text{ are also in A}\text{.P}$
$\begin{align} & \text{The common difference } \\ & \frac{1}{\alpha +\gamma }-\frac{1}{\beta +\gamma }=\frac{1}{\alpha +\beta }-\frac{1}{\alpha +\gamma } \\ & \frac{\beta +\gamma -\alpha -\gamma }{(\alpha +\gamma )(\beta +\gamma )}=\frac{\alpha +\gamma -\alpha -\beta }{(\alpha +\beta )(\alpha +\gamma )} \\ & \frac{\beta -\alpha }{(\alpha +\gamma )(\beta +\gamma )}=\frac{\gamma -\beta }{(\alpha +\beta )(\alpha +\gamma )} \\ & \frac{\beta -\alpha }{(\beta +\gamma )}=\frac{\gamma -\beta }{(\alpha +\beta )} \\ & (\beta -\alpha )(\alpha +\beta )=(\gamma -\beta )(\beta +\gamma ) \\ & {{\beta }^{2}}-{{\alpha }^{2}}={{\gamma }^{2}}-{{\beta }^{2}} \\ & 2{{\beta }^{2}}={{\gamma }^{2}}+{{\alpha }^{2}} \\ & {{\beta }^{2}}=\frac{{{\gamma }^{2}}+{{\alpha }^{2}}}{2}----(proved) \\ & {{\alpha }^{2}},{{\beta }^{2}},{{\gamma }^{2}}\text{ are also in A}\text{.P} \\\end{align}$