Explanation

(c)
$\because T_{6} = 2\ \Rightarrow a + 5d = 2$
Now, let $P = T_{1}T_{4}T_{5}$
$= a(a + 3d)(a + 4d)$
$= (2 - 5d)(2 - 2d)(2 - d)$
$= 2\{ 4 - 16d + 17d^{2} - 5d^{3}\}$
Now, $\frac{dp}{dd} = 2\left\{ - 16 + 34d - 15d^{2} \right\}$
Put $\frac{dp}{dd} = 0$ for maximum or minimum
$- 16 + 34d - 15d^{2} = 0$
$\Rightarrow \ \ d = \frac{2}{3}$ and $\frac{8}{5}$
Also, $\frac{d^{2}P}{dd^{2}} = 2\{ 34 - 30d\}$
$\left( \frac{d^{2}P}{dd^{2}} \right)_{d = 2/3} > 0$
Thus, $P$ is least.
Thus, the value of $d = 2/3.$