What will be the maximum sum of 44, 42, 40, ... ? A. 502 B. 504 C. 506 D. 500

GATE 2013 · General Aptitude · Arithmetic Series · easy

Answer: Maximum sum = 506

1. Find the last non-negative term: T_n = 44 + (n-1)*(-2) = 44 - 2n + 2 = 46 - 2n. Setting T_n >= 0: 46 - 2n >= 0 => n <= 23. So T_22 = 46 - 44 = 2, T_23 = 46 - 46 = 0, T_24 = -2.
2. Compute the maximum sum: Using n = 22, a = 44, T_22 = 2: S_22 = 22/2 * (44 + 2) = 11 * 46 = 506. Verify: S_23 = 23/2 * (44 + 0) = 23/2 * 44 = 23 * 22 = 506. Both confirm maximum sum = 506.