The sixth term of an AP is 2, and its common difference is greater than one. The value of the common difference of the progression so that the product of the first, fourth and fifth terms is greatest is

Let a be the first term and d the common difference of the given AP. It is given that: a +5d =2 and d >1. 
Let P be the product of the first, fourth and fifth terms of the given A.P. Then,

$\begin{array}{rl} & \mathrm{P}=\mathrm{a}(\mathrm{a}+3\mathrm{d})(\mathrm{a}+4\mathrm{d})\\ \Rightarrow & \mathrm{P}=(2-5\mathrm{d})(2-2\mathrm{d})(2-\mathrm{d})\\ \Rightarrow & \mathrm{P}=2(1-\mathrm{d})(2-\mathrm{d})(2-5\mathrm{d})\\ \Rightarrow & \mathrm{P}=2(-5{\mathrm{d}}^3+17{\mathrm{d}}^2-16\mathrm{d}+4)\end{array}$

We have to find the maximum value of P For maximum or minimum, we must have  

$\begin{array}{rl} & {\mathrm{P}}^{\mathrm{\prime }}\left(\mathrm{d}\right)=0\\ \Rightarrow & 2(-15{\mathrm{d}}^2+34\mathrm{d}-16)=0\\ \Rightarrow & 15{\mathrm{d}}^2-34\mathrm{d}+16=0\end{array}$

$\Rightarrow (3\mathrm{d}-2)(5\mathrm{d}-8)=0\Rightarrow \mathrm{d}=2/3\text{or}\mathrm{d}=8/5$

now, ${\mathrm{P}}^{\prime \prime }\left(\mathrm{d}\right)=2(-30\mathrm{d}+34)$

Clearly, ${\mathrm{P}}^{\prime \prime }\left(\mathrm{d}\right)<0 \text{for }\mathrm{d}=8/5.$So, P is maximum for $\mathrm{d}=8/5$