Let ${\alpha }_1,{\beta }_1$ are the roots of $x^2-6x+p=0$ and ${\alpha }_2,{\beta }_2$ are the roots of $x^2-54x+q=0.\text{ If }{\alpha }_1,{\beta }_1,{\alpha }_2,{\beta }_2$

form an increasing GP. Then, the value of $(q-p)$ is

Given, $,{\alpha }_1,{\beta }_1$ are the roots of the equation

$\begin{array}{r}\Rightarrow x^2-6x+p\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ \Rightarrow {\alpha }_1+{\beta }_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=6\\ {\alpha }_1\cdot {\beta }_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=p\end{array}$    (i)   (ii)

and $,{\alpha }_1,{\beta }_1$ are the roots of the equation

$\begin{array}{r}{x}^2-54x+q=0\\ {\alpha }_2+{\beta }_2=54\\ {\alpha }_2\cdot {\beta }_2=q\end{array}$   (iii)  (iv)

Since ${\alpha }_1,{\beta }_1,{\alpha }_2,{\beta }_2$ are in GP

$\therefore {\alpha }_1=a_1{\beta }_1=a{r}_1{\alpha }_2=a{r}^2,{\beta }_2=a{r}^3$

On substituting these values in Eqs. (i), (ii) (iii) and (iv), we get 

$\begin{array}{r}a+ar=6\\ \Rightarrow a(1+r)=6\\ \Rightarrow a^{2}r=p\\ \Rightarrow a{r}^2+a{r}^3=54\\ \Rightarrow a{r}^2(1+r)=54\\ \Rightarrow a^2{r}^5=q\end{array}$... (v) ... (vi) ... (vii)... (viii) 

On dividing Eq. (vii) by Eq. (v), we get 

$\begin{array}{r}{r}^2=\frac{54}{6}=9\\ \Rightarrow r=\pm 3\\ \Rightarrow r=3\end{array}$

From Eq. (v), we have

$\text{ When }r=3,\text{ then }a(1+3)=6$

$a=\frac{6}{4}=\frac{3}{2}$

$\begin{array}{l}{\alpha }_1=\frac{3}{2}\\ {\alpha }_2=a{r}^2=\frac{3}{2}(3{)}^2=\frac{27}{2}\\ {\beta }_1=ar=\frac{3}{2}\left(3\right)=\frac{9}{2}\\ {\beta }_2=a{r}^3=\frac{3}{2}(3{)}^3=\frac{81}{2}\\ q-\rho ={\alpha }_2{\beta }_2-{\alpha }_1{\beta }_1\left[\text{ from Eqs. (ii) and (iv) }\right]\\ =\frac{27}{2}\times \frac{81}{2}-\frac{3}{2}\times \frac{9}{2}=\frac{1}{4}[2187-27]\\ =\frac{1}{4}\left(2160\right)=540\end{array}$