Let  ${\mathrm{\alpha }}_1,{\mathrm{\beta }}_1$be the roots ${\mathrm{x}}^2-6\mathrm{x}+\mathrm{p}=0\text{ and }{\mathrm{\alpha }}_2,{\mathrm{\beta }}_2$ be the roots of ${\mathrm{x}}^2-54\mathrm{x}+\mathrm{q}=0.\text{ If }{\mathrm{\alpha }}_1,{\mathrm{\beta }}_1,{\mathrm{\alpha }}_2,{\mathrm{\beta }}_2$ from 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, ${\mathrm{\alpha }}_1+{\mathrm{\beta }}_1=6\Rightarrow \mathrm{A}(1+\mathrm{R})=6 \left(1\right)$
                 ${\mathrm{\alpha }}_1{\mathrm{\beta }}_1=\mathrm{p}\Rightarrow {\mathrm{A}}^2\mathrm{R}=\mathrm{p} \left(2\right)$
Also, ${\mathrm{\alpha }}_2+{\mathrm{\beta }}_2=54\Rightarrow {\mathrm{AR}}^2(1+\mathrm{R})=54 \left(3\right)$
            ${\mathrm{\alpha }}_2{\mathrm{\beta }}_2=\mathrm{q}\Rightarrow {\mathrm{A}}^2{\mathrm{R}}^5=\mathrm{q} \left(4\right)$
Now, on dividing Eq. (3) by Eq. (1), we get
$\frac{{\mathrm{AR}}^2(1+\mathrm{R})}{\mathrm{A}(1+\mathrm{R})}=\frac{54}{6}=9\text{ or }{\mathrm{R}}^2=9$
$\therefore$ R = 3 (as it is an increasing G.P.)
$\therefore$On putting R = 3 in Eq. (1), we get
$\mathrm{A}=\frac{6}{4}=\frac{3}{2}$
$\begin{array}{r}\therefore \mathrm{p}={\mathrm{A}}^2\mathrm{R}=\frac{9}{4}\times 3=\frac{27}{4}\text{ and }\\ \mathrm{q}={\mathrm{A}}^2{\mathrm{R}}^5=\frac{9}{4}\times 243=\frac{2187}{4}\end{array}$
Hence, $\mathrm{q}-\mathrm{p}=\frac{2187-27}{4}=\frac{2160}{4}=540$