Let A(a−d,a,a+d)a=12p=12(144−d2)q=12(144−D2)p−qp+q=519B(b−D,b,b+D)b=12
qp=1424=712144−(d2+6d+9)144−d2=7121008−7d2=−12d2−72d+16205d2+72d−612=0d=6D=9p−q=12(D2−d3)=12(81−36)=12(45)=540 option (4)
JEE Main 2025 — Mathematics Algebra
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 AP's in A and B respectively such that D=d+3,d>0. If p−qp+q=519, then p−q is equal to
Held on 4 Apr 2025 · Verified 6 Jul 2026.
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The number of elements in the set $S = \left\{(r, k) : k \in \mathbb{Z} \text{ and } {}^{36}C_{r+1} = \dfrac{6\left({}^{35}C_r\right)}{(k^2 - 3)}\right\}$, is :
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