Let $p$ and $q$ be the roots of the quadratic equation $ax^2+bx+c=0$, $a\ne0$. If $a$, $b$ and $c$ are in arithmetic progression and $\frac{1}{p}+\frac{1}{q}=4$, then the value of $(p-q)^2$ is

Using the formulas for the sum and product of roots of $ax^2+bx+c=0$:

$p+q=-\dfrac{b}{a}$ and $pq=\dfrac{c}{a}$

The given condition becomes

$\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{p+q}{pq}=\dfrac{-\frac{b}{a}}{\frac{c}{a}}=-\dfrac{b}{c}=4$

$\Rightarrow b=-4c$

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Since $a$, $b$, $c$ are in arithmetic progression, the middle term is the average of the other two:

$2b=a+c$

$\Rightarrow 2(-4c)=a+c$

$\Rightarrow a=-9c$

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Now put these back:

$p+q=-\dfrac{b}{a}=-\dfrac{-4c}{-9c}=-\dfrac{4}{9}$

$pq=\dfrac{c}{a}=\dfrac{c}{-9c}=-\dfrac{1}{9}$

Using the identity $(p-q)^2=(p+q)^2-4pq$:

$(p-q)^2=\left(-\dfrac{4}{9}\right)^2-4\left(-\dfrac{1}{9}\right)$

$=\dfrac{16}{81}+\dfrac{4}{9}$

$=\dfrac{16}{81}+\dfrac{36}{81}$

$=\dfrac{52}{81}$