JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The total number of ways to select 3 different dates from a month of 31 days is given by choosing 3 days out of 31: ^ 31 C_ 3 = 31 30 29 3 2 1 = 4495 Let the three selected dates be x, y, z such that x 2y = x + z For y to be an integer, the sum x + z must be an even number. This is only possible if both x and z are either odd or even. Once x and z are chosen, y is uniquely determined as their exact average. In a month of 31 days, the number of odd dates (1, 3, 5, , 31) is 16, and the number of even dates (2, 4, 6, , 30) is 15. The number of ways to choose x and z such that both are odd is: ^ 16 C_ 2 = 16 15 2 = 120 The number of ways to choose x and z such that both are even is: ^ 15 C_ 2 = 15 14 2 = 105 The total number of favorable selections that form an increasing A.P. is: 120 + 105 = 225 The probability that the selected dates are in an increasing A.P. is: P = 225 4495 Dividing the numerator and the denominator by 5, we get: P = 45 899 Since 45 = 3^2 5 and 899 = 29 31, they share no common factors, meaning (45, 899) = 1. Thus, a = 45 and b = 899. Finally, a + b = 45 + 899 = 944. Answer: 944