If the roots of the cubic equation ${\mathrm{ax}}^2+{\mathrm{dx}}^2+\mathrm{cx}+\mathrm{d}$ are in GP ,then

Let$\frac{\mathrm{\alpha }}{\mathrm{r}},\mathrm{\alpha }$ and $\mathrm{\alpha }$ r be the roots of the equation

${\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}=0$

$\therefore \text{ Sum of the roots, }\frac{\mathrm{\alpha }}{\mathrm{r}}+\mathrm{\alpha }+\mathrm{\alpha r}=-\frac{\mathrm{b}}{\mathrm{a}}$

$\Rightarrow \mathrm{\alpha }\left(\frac{1}{\mathrm{r}}+1+\mathrm{r}\right)=-\frac{\mathrm{b}}{\mathrm{a}} ....\left(\mathrm{i}\right)$

Sum  of product  of the  roots  taken two  at a time,

$\begin{array}{r}\frac{\mathrm{\alpha }}{\mathrm{r}}\cdot \mathrm{\alpha }+\mathrm{\alpha }\cdot \mathrm{\alpha r}+\mathrm{\alpha r}\cdot \frac{\mathrm{\alpha }}{\mathrm{r}}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{\mathrm{c}}{\mathrm{a}}\\ \Rightarrow {\mathrm{\alpha }}^2\left(\frac{1}{\mathrm{r}}+\mathrm{r}+1\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{\mathrm{c}}{\mathrm{a}}......\left(\mathrm{ii}\right)\end{array}$

Product  of the  roots, $\frac{\mathrm{\alpha }}{\mathrm{r}}\cdot \mathrm{\alpha }\cdot \mathrm{\alpha r}=-\frac{\mathrm{d}}{\mathrm{a}}$

$\Rightarrow {\mathrm{\alpha }}^3=-\frac{\mathrm{d}}{\mathrm{a}} .....\left(\mathrm{iii}\right)$

On  dividing  Eqs. (ii)  by (i), we  get

$\frac{{\mathrm{\alpha }}^2\left(\frac{1}{\mathrm{r}}+\mathrm{r}+1\right)}{\mathrm{\alpha }\left(\frac{1}{\mathrm{r}}+1+\mathrm{r}\right)}=\frac{\frac{\mathrm{c}}{\mathrm{a}}}{-\frac{\mathrm{b}}{\mathrm{a}}}$

$\Rightarrow \mathrm{\alpha }=-\frac{\mathrm{c}}{\mathrm{b}} ...\left(\mathrm{iv}\right)$

$\begin{array}{l}{\left(-\frac{\mathrm{c}}{\mathrm{b}}\right)}^3=-\frac{\mathrm{d}}{\mathrm{a}}\Rightarrow -\frac{{\mathrm{c}}^3}{{\mathrm{b}}^3}=-\frac{\mathrm{d}}{\mathrm{a}}\Rightarrow \frac{{\mathrm{c}}^3}{{\mathrm{b}}^3}=\frac{\mathrm{d}}{\mathrm{a}}\\ \Rightarrow {\mathrm{c}}^3\cdot \mathrm{a}=\mathrm{d}\cdot {\mathrm{b}}^3\end{array}$