Let a , b be two non-zero real numbers. If p and r are the roots of the equation x 2 - 8 a x + 2 a = 0 and q…

Since $p, r$ are roots of $x^2-8ax+2a=0$ and

$q,s$ are the roots of $x^2+12bx+6b=0$ 

So the equation $2a{x}^2-8ax+1=0$ has roots $\frac{1}{p},\frac{1}{r}$ and

$6b{x}^2+12bx+1=0$ has roots $\frac{1}{q},\frac{1}{s}$.

Given that $\frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{s}$ are in A.P.

Let $\frac{1}{p}=\alpha -3\beta ,\frac{1}{q}=\alpha -\beta ,\frac{1}{r}=\alpha +\beta$ and $\frac{1}{s}=\alpha +3\beta$

i.e. $\frac{1}{p}+\frac{1}{r}=2\alpha -2\beta =4$ and

$\frac{1}{q}+\frac{1}{s}=2\alpha +2\beta =-2$

or $\alpha =\frac{1}{2}, \beta =-\frac{3}{2}$

i.e. $\frac{1}{p}=5, \frac{1}{q}=2, \frac{1}{r}=-1, \frac{1}{s}=-4$

Now 

$\frac{1}{p}\times \frac{1}{r}=\frac{1}{2a}=-5\Rightarrow \frac{1}{a}=-10$ and $\frac{1}{q}\times \frac{1}{s}=\frac{1}{6b}=-8\Rightarrow \frac{1}{b}=-48$

Hence, $a^{-1}-b^{-1}=-10+48=38$