JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

Given a, b, c are in arithmetic progression, we have 2b = a + c. Let the integer roots of the equation ax^2 + bx + c = 0 be and . Sum of the roots is + = - b a and product of the roots is = c a . Dividing the arithmetic progression condition by a, we get: 2b a = 1 + c a Substituting the sum and product of the roots: -2( + ) = 1 + + 2 + 2 + 1 = 0 Adding 3 to both sides to factorize: + 2 + 2 + 4 = 3 ( + 2)( + 2) = 3 Since and are integers, ( + 2) and ( + 2) must be integer factors of 3. The possible pairs are (1, 3), (3, 1), (-1, -3), and (-3, -1). If ( + 2, + 2) = (1, 3) or (3, 1), the roots are -1 and 1. The sum of the roots is 0, which implies b = 0. This is rejected because b is a positive integer. If ( + 2, + 2) = (-1, -3) or (-3, -1), the roots are -3 and -5. The sum of the roots is -8, which implies - b a = -8 b = 8a. The product of the roots is 15, which implies c a = 15 c = 15a. Since a is a positive integer, b = 8a and c = 15a are also positive integers. The roots of the equation are always -3 and -5. Evaluating the given statements: c - b = 15a - 8a = 7a, which is an integer multiple of a. The roots are -3 and -5, both of which are odd integers. If c = 15, then 15a = 15 a = 1. Consequently, b = 8(1) = 8. Thus, ab = 1 8 = 8. If b = 8, then 8a = 8 a = 1. The roots are -3 and -5, so x = 3 is not a root. Answer: c - b is an integer multiple of a; Both the roots of the equation ax^2 + bx + c = 0 are odd integers; If c = 15, then ab = 8