Let a,b be two non-zero real numbers if p and r are the roots of the equation ${\mathrm{x}}^2-8\mathrm{ax}+2\mathrm{a}=0$ and q and s are the roots of the equation ${\mathrm{x}}^2+12\mathrm{bx}+6\mathrm{a}=0$such that $\frac{1}{\mathrm{p}},\frac{1}{\mathrm{q}},\frac{1}{\mathrm{r}},\frac{1}{\mathrm{s}}\text{ are in A.P then }\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}=$

$$\begin{gathered} \begin{array}{l}{\mathrm{x}}^2-8\mathrm{ax}+2\mathrm{a}=0\text{ Roots are }\mathrm{p},\mathrm{r} \\ \mathrm{p}+\mathrm{r}=8\mathrm{a}, \mathrm{p}r=2\mathrm{a} \\ \Rightarrow \frac{\mathrm{p}+\mathrm{r}}{\mathrm{pr}}=\frac{8\mathrm{a}}{2\mathrm{a}}\Rightarrow \frac{1}{\mathrm{r}}+\frac{1}{\mathrm{p}}=4\Rightarrow \frac{2}{\mathrm{q}}=4\\ \Rightarrow \mathrm{q}=\frac{1}{2} \\ {\mathrm{x}}^2+12\mathrm{bx}+6\mathrm{a}=0\to \text{ roots are }\mathrm{q},s \\ q+\mathrm{s}=-12\mathrm{b}, \mathrm{qs}=6\mathrm{b} \\ \Rightarrow \frac{\mathrm{q}+\mathrm{s}}{\mathrm{qs}}=\frac{-12b}{6b}=-2\\ \Rightarrow \frac{1}{\mathrm{s}}+\frac{1}{\mathrm{q}}=-2\Rightarrow \frac{2}{\mathrm{r}}=-2\Rightarrow \mathrm{r}=-1 \\ \mathrm{p}=\frac{1}{5} \\ \mathrm{s}=\frac{-1}{4} \\ \frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}=\frac{2}{\mathrm{pr}}-\frac{6}{\mathrm{qs}} \\ =\frac{2}{\frac{1}{5}(-1)}-\frac{6}{\frac{1}{2}\left(\frac{-1}{4}\right)}=-10+48=38\end{array} \end{gathered}$$