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 $\mathrm{d}=8/5$