JEE Main 2025 — 4 April, Evening Shift — Question 20

Let the terms in $A$ be $a_{1}-d, a_{1}, a_{1}+d$ and in $B$ be $a_{2}-D, a_{2}, a_{2}+D$

Now $3 a_{1}=36$

$\Rightarrow a_{1}=12$ and $3 a_{2}=36$

$\Rightarrow a_{2}=12$

Now $(12-d)(12)(12+d)=p$ and $(12-D)(12)(12+D)=q$

Also $\frac{p+q}{p-q}=\frac{19}{5}$

$\Rightarrow 12 q=7 p$

$\Rightarrow 12(12-D)(12)(12+D)=7(12-d)(12)(12+d)$

$\Rightarrow 12(9-d)(12)(15-d)=7(12-d)(12)(12+d)$

$\Rightarrow 12\left(135-d^{2}-6 d\right)=7\left(144-d^{2}\right)$

$\Rightarrow d=6, D=9$

$p=6 \times 12 \times 18=1296$

$q=756$

$p-q=540$