JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let = a, = ar, = ar^2, = ar^3 be the terms of the G.P. From the given quadratic equations, the sum of the roots are: + = 1 a(1 + r) = 1 + = 4 ar^2(1 + r) = 4 Dividing the second equation by the first equation: ar^2(1 + r) a(1 + r) = 4 1 r^2 = 4 r = 2 or r = -2 If r = 2, then a(1 + 2) = 1 a = 1 3 . The product of the roots of the first equation is p = = a^2 r = ( 1 3 )^2 2 = 2 9 . Since p Z , r = 2 is rejected. If r = -2, then a(1 - 2) = 1 a = -1. The product of the roots of the first equation is p = = a^2 r = (-1)^2 (-2) = -2. The product of the roots of the second equation is q = = (ar^2)(ar^3) = a^2 r^5 = (-1)^2 (-2)^5 = -32. Both p and q are integers, which satisfies the given condition. Therefore, |p + q| = |-2 - 32| = |-34| = 34. Answer: 34