Let $\mathrm{\alpha }, \mathrm{\beta }$ be the roots of ${\mathrm{x}}^2-\mathrm{x}+\mathrm{p}=0$ and $\mathrm{\gamma }, \mathrm{\delta }$ be the roots of ${\mathrm{x}}^2-4\mathrm{x}+\mathrm{q}=0$. If $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }, \mathrm{\delta }$ are in G.P., then integral values of p, q are respectively

Let r be the common ratio of the G.P. $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }, \mathrm{\delta }$ then $\mathrm{\beta }=\mathrm{\alpha r}, \mathrm{\gamma }={\mathrm{\alpha r}}^2 \mathrm{and} \mathrm{\delta }={\mathrm{\alpha r}}^3$

$\therefore \mathrm{\alpha }+\mathrm{\beta }=1\Rightarrow \mathrm{\alpha }+\mathrm{\alpha r}=1\Rightarrow \mathrm{\alpha }(1+\mathrm{r})=1 ...\left(\mathrm{i}\right)$

$\mathrm{\alpha \beta }=\mathrm{p}\Rightarrow \mathrm{\alpha }\left(\mathrm{\alpha r}\right)=\mathrm{p}\Rightarrow {\mathrm{\alpha }}^2{\mathrm{r}}^2=\mathrm{P} ...\left(\mathrm{ii}\right)$

$\mathrm{\gamma }+\mathrm{\delta }=4\Rightarrow {\mathrm{\alpha r}}^2+{\mathrm{ar}}^3=4\Rightarrow {\mathrm{\alpha r}}^2(1+\mathrm{r})=4 ....\left(\mathrm{iii}\right)$

and $\mathrm{\gamma \delta }=\mathrm{q}\Rightarrow {\mathrm{\alpha r}}^2.{\mathrm{\alpha r}}^3=\mathrm{q}\Rightarrow {\mathrm{\alpha }}^2{\mathrm{r}}^5=\mathrm{q} ...\left(\mathrm{iv}\right)$

Dividing (iii) by (i), we get, ${\mathrm{r}}^2=4\Rightarrow \mathrm{r}=\pm 2$

If we take $\mathrm{r}=2$, then $\mathrm{\alpha }$ is not integral, so we take $\mathrm{r}=-2,$

Substituting $\mathrm{r}=-2 \mathrm{in} \left(\mathrm{i}\right), \mathrm{we} \mathrm{get} \mathrm{\alpha }=-1$

Now, from (ii), we have $\mathrm{p}={\mathrm{\alpha }}^2\mathrm{r}={(-1)}^2(-2)=-2$

and from (iv), we have $\mathrm{q}={\mathrm{\alpha }}^2{\mathrm{r}}^5={(-1)}^2{(-2)}^5=-32$

$\Rightarrow (\mathrm{p},\mathrm{q})=(-2, -32)$