Let $\alpha ,\beta$ be the roots of the equation $p{x}^2+qx+r=0$, $p\ne 0$. If $p,q,r$ are in AP and $\frac{1}{\alpha }+\frac{1}{\beta }=4$, the value of $\left|\begin{array}{c}\alpha -\beta \end{array}\right|$ is 

$\because \frac{1}{\alpha }+\frac{1}{\beta }=4\Rightarrow \frac{\alpha +\beta }{\alpha \beta }=4$

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{-q}{p}}}{{\displaystyle \frac{r}{p}}}=4 \\ \Rightarrow q=-4r....\left(i\right) \end{gathered}$$

Also, given $p,q,r$ are in AP

$\therefore 2q=p+r$

$\Rightarrow p=-9r$    [From Eq. (i)]$....\left(ii\right)$

Now, $\left|\begin{array}{c}\alpha -\beta \end{array}\right|=\frac{\sqrt{D}}{\left|\begin{array}{c}a\end{array}\right|}$$[\because a{x}^2+bx+c=0, \alpha -\beta =\frac{\sqrt{D}}{a}]$

$=\frac{\sqrt{(q^2-4pr)}}{\left|\begin{array}{c}p\end{array}\right|}$

$=\frac{\sqrt{16r^2+36r^2}}{9\left|\begin{array}{c}r\end{array}\right|}=\frac{\sqrt{52}\left|\begin{array}{c}r\end{array}\right|}{9\left|\begin{array}{c}r\end{array}\right|}$   [From Eqs (i) and (ii)]

$=\frac{2\sqrt{13}}{9}$