The `6th` term of an `AP` is equal to `2`. The value of the common difference of the `AP` which makes the product `a_1a_4a_5` least is given by

Options

A. `(8)/(5)`
B. `(5)/(4)`
C. `(2)/(3)`
D. None of these

Answer

C

Explanation

`T_(6)=2`
Let d be common difference of AP and a be the first term of AP.
`T_(6)=2`
`implies a+5d=2 " " "….(i)"`
Let `A=a_(1)a_(4)a_(5)`
`Aa(a+3d)(a+4d)`
[using `T_(n)=a+(n-1)d` and from Eq. (i)`a=2-5d`]
`A=(2-5d)(2-2d)(2-d)`
`A=8-32d+34d^(2)=10d^(3)`
For max and min values of `A,(dA)/(dd)=0`
`-30d^(2)+68d-32=0 " " implies 15d^(2)-34d+16=0`
`15d^(2)-(24d+19d)+16=0`
`15d^(2)-24d-10d+16=0`
`3d(5d-8)-2(5d-8)=0`
`(5d-8(3d-2)=0`
`d=(8)/(5)` or `d=(2)/(3)`
`d=(2)/(3),(d^(2)A)/(dd^(2))gt0`
Ais least for `d=(2)/(3)`.