Explanation

(d)
Since, $a_{1},\ a_{2},\ a_{3},\ldots,\ a_{10}\$be in AP.
$\therefore\ a_{10} = a_{1} + 9d$
$\Rightarrow 3 = a_{1} + 9d$
$\Rightarrow 3 = 2 + 9d$
$\Rightarrow d = \frac{1}{9}$
Now, $a_{4} = a_{1} + 3d$
$\Rightarrow a_{4} = 2 + 3\left( \frac{1}{9} \right) = 2 + \frac{1}{3} = \frac{7}{3}$
Again, $h_{1},\ h_{1},\ h_{3},\ldots,\ h_{10}$ be in HP.
$\Rightarrow \frac{1}{h_{1}},\frac{1}{h_{2}},\frac{1}{h_{3}},\ldots,\frac{1}{h_{10}}$ be in AP
Since, $h_{1} = 2,\ h_{10} = 3$ (given)
$\therefore\ \frac{1}{h_{10}} = \frac{1}{h_{1}} + 9d_{1}$
$\Rightarrow \frac{1}{3} = \frac{1}{2} + 9d_{1}$
$\Rightarrow \frac{1}{3} - \frac{1}{2} = 9d_{1}$
$\Rightarrow - \frac{1}{6} = 9d_{1}$
$\Rightarrow d_{1} = - \frac{1}{54}$
Now, $\frac{1}{h_{7}} = \frac{1}{2} + \frac{6 \times 1}{- 54}\ \ \Rightarrow \frac{1}{h_{7}} = \frac{1}{2} - \frac{1}{9}$
$\Rightarrow \frac{1}{h_{7}} = \frac{9 - 2}{18}\ \Rightarrow \ h_{7} = \frac{18}{7}$
$\therefore a_{4}h_{7} = \frac{7}{3} \times \frac{18}{7} = 6$