Let a 1 , a 2 , a 3 . . . be an A . P . with a 6 = 2 . Then, the common difference of this A . P . , which maximise the product a 1 · a 4 · a 5 , is :

Step-by-step solution

Let, the first term of A . P . be a and common difference d , then we know that the n t h term of the A . P . is a + n - 1 d . Then a 6 = a + 5 d = 2 a = 2 - 5 d       . . . i Let ∆ = a 1 · a 4 · a 5 ⇒ ∆ = a · ( a + 3 d ) · ( a + 4 d ) Put the value of a from equation i , ⇒ ∆ = 2 - 5 d · ( 2 - 5 d + 3 d ) · ( 2 - 5 d + 4 d ) ⇒ ∆ = 2 - 5 d · ( 2 - 2 d ) · ( 2 - d ) ⇒ ∆ = - 2 ( 5 d 3 - 17 d 2 + 16 d - 4 ) For