Let ${\mathrm{\alpha }}_1, {\mathrm{\beta }}_1$ be the roots of ${\mathrm{x}}^2-6\mathrm{x}+\mathrm{p}=0$ and ${\mathrm{\alpha }}_2, {\mathrm{\beta }}_2$ be the roots of ${\mathrm{x}}^2-54\mathrm{x}+\mathrm{q}=0.\text{ If }{\mathrm{\alpha }}_1, {\mathrm{\beta }}_1, {\mathrm{\alpha }}_2, {\mathrm{\beta }}_2$ form an increasing G.P., then the value of (q - p) is _________.

Let ${\mathrm{\alpha }}_1=\mathrm{A}, {\mathrm{\beta }}_1=\mathrm{AR}, {\mathrm{\alpha }}_2={\mathrm{AR}}^2, {\mathrm{\beta }}_2={\mathrm{AR}}^3$

we have ${\mathrm{\alpha }}_1+{\mathrm{\beta }}_1=6\Rightarrow \mathrm{A}(1+\mathrm{R})=6$           ….. (1)

${\mathrm{\alpha }}_1{\mathrm{\beta }}_1=\mathrm{p}\Rightarrow {\mathrm{A}}^2\mathrm{R}=\mathrm{p}$                                   …… (2)

Also ${\mathrm{\alpha }}_2+{\mathrm{\beta }}_2=54\Rightarrow {\mathrm{AR}}^2(1+\mathrm{R})=54$        ……. (3)

${\mathrm{\alpha }}_2{\mathrm{\beta }}_2=\mathrm{q}\Rightarrow {\mathrm{A}}^2{\mathrm{R}}^5=\mathrm{q}$                                 …….. (4)

Now, on dividing Eq. (3) by Eq. (1), we get

$\frac{{\mathrm{AR}}^2(1+\mathrm{R})}{\mathrm{A}(1+\mathrm{R})}=\frac{54}{6}=9\text{ or }{\mathrm{R}}^2=9$

$\therefore \mathrm{R}=3$ (as it is an increasing G.P.)

$\therefore$ On putting R - 3 in Eq. (1), we get

$\begin{array}{c}\mathrm{A}=\frac{6}{4}=\frac{3}{2}\\ \therefore \mathrm{p}={\mathrm{A}}^2\mathrm{R}=\frac{9}{4}\times 3=\frac{27}{4}\end{array}$

and $\mathrm{q}={\mathrm{A}}^2{\mathrm{R}}^5=\frac{9}{4}\times 243=\frac{2187}{4}$

Hence, $\mathrm{q}-\mathrm{p}=\frac{2187-27}{4}=\frac{2160}{4}=540$