In a harmonic progression t 1 , t 2 , t 3 , … … … . . , it is given that t 5 = 20 and t 6 = 50. If S n denotes the sum of first n terms of this, then the value of n for which S n is maximum is

Step-by-step solution

Let the first term be a and the common difference be d , then 20 = 1 a + 4 d and 50 = 1 a + 5 d On solving, we get, d = - 3 100 and a = 17 100 Hence, t r = 100 20 - 3 r > 0 ⇒ r < 20 3 ⇒ r m a x = 6 So, sum is maximum for n = 6, as after this terms are negative