The 4 th term of GP is 500 and its common ratio is 1 m , m ∈ N . Let S n denote the sum of the first n terms of this GP. If S 6 > S 5 + 1 and S 7 < S 6 + 1 2 , then the number of possible values of m is ______

Step-by-step solution

Given: T 4 = 500 Let a = first term r = common ratio = 1 m ,   m ∈ N So, a r 3 = 500 ⇒ a m 3 = 500 Now, S n - S n - 1 = a r n - 1 S 6 > S 5 + 1 ⇒ S 6 - S 5 > 1 ⇒ a m 5 > 1 ⇒ a m 3 × 1 m 2 > 1 ⇒ 500 m 2 > 1 ⇒ m 2 < 500       . . . 1 And, S 7 - S 6 < 1 2 ⇒ a m 6 < 1 2 ⇒ a m 3 × 2 m 3 < 1 ⇒ 500 × 2 m 3 < 1 ⇒ m 3 > 10 3 ⇒ m > 10       . . . . 2 From 1