For p, q ∈ R , consider the real valued function f(x) = (x - p) ^ 2...

Given,

$f\left(x\right) = 0 ⇒ {\left(x - p\right)}^{2} - q = 0$

So, roots are $p + \sqrt{q}, p - \sqrt{q}$

Now absolute difference between roots will be $2\sqrt{q}$.

Now given ${a}_{1}, {a}_{2}, {a}_{3}, {a}_{4}$ are in $A . P$ and its mean is $p$

Now let ${a}_{1}, {a}_{2}, {a}_{3}, {a}_{4}$ be ${a}_{1} = p - 3d, {a}_{2} = p - d, {a}_{3} = p + d \text{\text{ and } }$ {a}_{4} = p + 3d

Now given \left|f\left({a}_{i}\right)\right| = 500$</p><p>So, \left|f\left({a}_{4}\right)\right| = 500$

$⇒ \left|{\left({a}_{4} - p\right)}^{2} - q\right| = 500$

$⇒ {\left({a}_{4} - p\right)}^{2} - q = 500$

$⇒ 9{d}^{2} - q = 500 . . . . \left(1\right)$

And using \left|f\left({a}_{i}\right)\right| = 500 \forall i = 1, 2, 3, 4$</p><p>We get${\left|f\left({a}_{4}\right)\right|}^{2} = {\left|f\left({a}_{3}\right)\right|}^{2}$</p><p>$⇒ {\left({\left({a}_{4} - p\right)}^{2} - q\right)}^{2} = {\left({\left({a}_{3} - p\right)}^{2} - q\right)}^{2}$</p><p>$⇒ 9{d}^{2} - q + {d}^{2} - q = 0$</p><p>So,$2q = 10{d}^{2} ⇒ q = 5{d}^{2}$</p><p>$⇒ {d}^{2} = \dfrac{q}{5}$</p><p>From equation \left(1\right)$ we get,

$9\left(\dfrac{q}{5}\right) - q = 500$

$⇒ \dfrac{4q}{5} = 500$

$⇒ q = \dfrac{500 \times 5}{4}$

Now absolute difference is $2\sqrt{q} = 2 \times \sqrt{\dfrac{500 \times 5}{4}} = 2 \times \dfrac{50}{2} = 50$