Let a 1 , a 2 , a 3 , … … . be an A.P. If a 7 = 3 , the product a 1 a 4 is minimum and the sum of its first n terms is zero then n ! - 4 a n n + 2 is equal to

Step-by-step solution

We know the n t h term of an A.P. is given by, a n = a + n - 1 d Given, a 7 = 3 ⇒ a + 6 d = 3 ⇒   a = 3 - 6 d And, a 1 a 4 = a a + 3 d = 3 - 6 d 3 - 3 d = 18 d 2 - 27 d + 9 Given product a 1 a 4 is minimum then, Let f ( d ) = 18 d 2 - 27 d + 9 f ' ( d ) = 36 d - 27 Product to be minimum, f ' d = 0 ⇒ 36 d - 27 = 0 ⇒ d = 27 36 = 3 4 So, a = 3 - 9 2 = - 3 2 Given, S n = 0 S n = n 2 2 a + n - 1 d = 0 - 3 + n - 1 3 4 = 0 ⇒   n = 5 Now n ! - 4 a n ( n + 2 ) = 5 ! - 4 a 35 = 120