For $p,q\in R,$ consider the real valued function $f\left(x\right)={(x-p)}^2-q, x\in R and q>0. Let a_1, a_2, a_3 and a_4$be in an arithmetic progression with mean p and positive common difference.

$If \left|f\left(a_i\right)\right|=500 for all i=1, 2, 3, 4,$ then the absolute difference between the roots of f(x)=0 is 

$$\begin{gathered} f\left(x\right)=0\Rightarrow {\left(x-p\right)}^2-q=0 \\ Roots are p+\sqrt{q}, p-\sqrt{q} absolute difference \\ between roots 2\sqrt{q}. \\ Now, \left|f\left(a_i\right)\right|=500 \\ Let a_1, a_2, a_3,a_4 are a, a+d, a+2d, a+3d \\ p=\frac{4a+6d}{4} \\ \left|f\left(a_1\right)\right|=500\Rightarrow \left|{\left(a-p\right)}^2-q\right|=500 \\ \left|{\left(a_1-p\right)}^2-q\right|=500 \\ \Rightarrow {\left(a_1-p\right)}^2-q=500 \\ \Rightarrow \frac{9}{4}{d}^2-q=500 \_\_\_\_\_\_ \left(1\right) \\ and {\left|f\left(a_1\right)\right|}^2={\left|f\left(a_2\right)\right|}^2 \\ {\left({\left(a_1-p\right)}^2-q\right)}^2={\left({\left(a_2-p\right)}^2-q\right)}^2 \\ \Rightarrow \left({\left(a_1-p\right)}^2-{\left(a_2-p\right)}^2\right)\left({\left(a_1-p\right)}^2-q+{\left(a_2-p\right)}^2-q\right)=0 \\ \Rightarrow \frac{9}{4}{d}^2-q+\frac{d^2}{4}-q=0 \\ 2q=\frac{10d^2}{4}\Rightarrow q=\frac{5d^2}{4} \\ \Rightarrow d^2=\frac{4q}{5} \\ From equation \left(1\right) \frac{9}{4}·\frac{4.q}{5}-q=500 \\ \frac{4q}{5}=500 \\ \frac{4q}{5}=500 \\ and 2\sqrt{q}=2\times \frac{50}{2}=50 \end{gathered}$$