If the sum of the first 100 terms of an arithmetic progression is - 1 and the sum of the even terms lying in the first 100 terms is 1 , then the 100 t h term of the arithmetic progression is

Step-by-step solution

Given, S 100 = - 1 ⇒ 100 2 2 a + 99 d = - 1 ⇒ 50 2 a + 99 d = - 1 Also, a 2 + a 4 + a 6 + . . . . . . + a 100 = 1 ⇒ 50 2 2 a + d + 49 2 d = 1 ⇒ 50 a + 50 d = 1 On solving, we get, d = 3 50 and a = - 149 50 a 100 = a + 99 d = 74 25