JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 18)
Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and $q$ respectively. Let d and D be the common differences of $\mathrm{AP}^{\prime} \mathrm{s}$ in
$A$ and $B$ respectively such that $D=d+3, d>0$. If $\frac{p+q}{p-q}=\frac{19}{5}$, then $\mathrm{p}-\mathrm{q}$ is equal to
540
450
600
630
Explanation
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 $3 a_1=36$
$$\Rightarrow \quad a_1=12$$
and $3 a_2=36$
$$\Rightarrow \quad a_2=12$$
Now $(12-d)(12)(12+d)=p$
and $(12-D)(12)(12+D)=q$
Also $\frac{p+q}{p-q}=\frac{19}{5}$
$$\begin{aligned} & \Rightarrow 12 q=7 p \\ & \Rightarrow 12(12-D)(12)(12+D)=7(12-d)(12)(12+\mathrm{d}) \\ & \Rightarrow 12(9-d)(12)(15-d)=7(12-d)(12)(12+\mathrm{d}) \\ & \Rightarrow 12\left(135-d^2-6 d\right)=7\left(144-d^2\right) \\ & \Rightarrow d=6, D=9 \\ & p=6 \times 12 \times 18=1296 \\ & q=756 \\ & p-q=540 \end{aligned}$$
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