JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given series is = 3+4+8+9+13+14+ up to 40 terms. We can split the series into two arithmetic progressions, each containing 20 terms: S_1 = 3 + 8 + 13 + up to 20 terms S_2 = 4 + 9 + 14 + up to 20 terms Using the sum formula for an AP, S_n = n 2 [2a + (n-1)d]: S_1 = 20 2 [2(3) + (20-1)5] = 10[6 + 95] = 1010 S_2 = 20 2 [2(4) + (20-1)5] = 10[8 + 95] = 1030 Therefore, = S_1 + S_2 = 1010 + 1030 = 2040. The exponent in the given expression is 1020 = 2040 1020 = 2. Thus, ( )^2 = ^2 is a root of the equation x^2+x-2=0. Solving x^2+x-2=0 gives (x+2)(x-1)=0 x = -2 or x = 1. Since ^2 0, we must have ^2 = 1. Given (0, 2 ), = 1 = 4 . We need to find the value of ^2 + 3 ^2 . Substituting = 4 : ^2 ( 4 ) + 3 ^2 ( 4 ) = ( 1 2 )^2 + 3 ( 1 2 )^2 = 1 2 + 3 2 = 2. Answer: 2