If S n = n 2 a + n 4 n - 1 d is the sum of the first n terms of an arithmetic progression, then the common difference is

Step-by-step solution

S 1 = a = t 1 S 2 = 4 a + d 2 = t 1 + t 2 ⇒ t 2 = S 2 - t 1 = 4 a + d 2 - a = 3 a + d 2 ⇒ common difference = t 2 - t 1 = 3 a + d 2 - a = 2 a + d 2