A man starts walking and covers 50 m in 1 minute, 90 m in 2 minutes, 130 m in 3 minutes and so on. Find the total distance covered by the man in 15 minutes. If the speed of the man is the same in the first and last minute and constant in the rest of the time, then find the distance covered by the person in 15 minutes.

The distances form an AP: 50, 90, 130, ... with first term a = 50 and common difference d = 40. This is distance covered BY THE END of each minute. Sum of 15 terms = 15/2 × [2(50) + (15-1)(40)] = 15/2 × [100 + 560] = 15/2 × 660 = 4950 m. If the problem asks for distance covered IN 15 minutes (not total), we need the 15th term alone: 50 + 14×40 = 610 m. The closest option is B (620 m). Options A (520), C (720), and D (750) don't match.