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 :

Let, the first term of $A.P.$ be $a$ and common difference $d,$ then we know that the $n^{th}$term of the $A.P.$ is $a+\left(n-1\right)d.$

Then $a_6=a+5d=2$

$a=2-5d ...\left(i\right)$

Let $∆=a_1·a_4·a_5$

$\Rightarrow ∆=a·(a+3d)·(a+4d)$

Put the value of $a$ from equation $\left(i\right),$

$\Rightarrow ∆=\left(2-5d\right)·(2-5d+3d)·(2-5d+4d)$

$\Rightarrow ∆=\left(2-5d\right)·(2-2d)·(2-d)$

$$$\Rightarrow ∆=-2(5d^3-17d^2+16d-4)$

For finding the maximum or minimum value of $∆$

$\Rightarrow \frac{d∆}{dd}=-2(15d^2-34d+16)$

$\Rightarrow \frac{d∆}{dd}=-2\left(5d-8\right)(3d-2)$

Now, $\frac{d∆}{dd}=0$

$\Rightarrow -2\left(5d-8\right)(3d-2)=0$

$\Rightarrow d=\frac{8}{5}$ or $d=\frac{2}{3}$

And, $\frac{d^2∆}{d{d}^2}=-2(30d-34)$

At $d=\frac{8}{5}, \frac{d^2∆}{d{d}^2}=-2\left(30\times \frac{8}{5}-34\right)=-28<0$ and at $d=\frac{2}{3}, \frac{d^2∆}{d{d}^2}=-2\left(30\times \frac{2}{3}-34\right)=28>0$

We know that if at any point the first derivative of a function is zero and the second derivative is negative, then it is the point of maxima of the function.

Hence, it is clear that $∆$ is maximum when $d=\frac{8}{5}.$