The sum of the roots of the equation is equal to the sum of squares of their reciprocals. Find whether $b{c}^2,c{a}^2$ and ab² are in AP, GP or HP?

Let $\alpha$ and $\beta$ be its roots.  It is given that $\alpha +\beta =\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$

$\begin{array}{l}\Rightarrow \alpha +\beta =\frac{{\alpha }^2+{\beta }^2}{(\alpha \beta {)}^2}\\ \Rightarrow \alpha +\beta =\frac{(\alpha +\beta {)}^2-2\alpha \beta }{(\alpha \beta {)}^2}\\ \Rightarrow -\frac{b}{a}=\frac{\frac{b^2}{a^2}-2\frac{c}{a}}{c^2}=\frac{b^2-2ac}{c^2}\\ \Rightarrow -\frac{b}{a}=\frac{b^2-2ac}{c^2}\\ \Rightarrow a{b}^2-2a^{2}c=-b{c}^2\\ \Rightarrow a{b}^2+b{c}^2=2a^{2}c\end{array}$

$\Rightarrow a{b}^2,a^{2}c,b{c}^2$  are in AP