(a): We have,\n $a_{6} = 2 \Rightarrow a + 5d = 2 \Rightarrow d = \frac{2 - a}{5}$\\\n Let $a_{1}a_{4}a_{5} = \lambda$\n\end{enumerate}\n\nLet $a_{1}a_{4}a_{5} = \lambda$\\\n$\Rightarrow \lambda = a(a + 3d)(a + 4d)$\n\n$= a\left( a^{2} + 7ad + 12d^{2} \right) = a^{3} + 7a^{2}d + 12ad^{2}$\n\n$= a^{3} + 7a^{2}\left( \frac{2 - a}{5} \right) + 12a\left( \frac{2 - a}{5} \right)^{2}\ \left\lbrack \because d = \frac{2 - a}{5} \right\rbrack\n$\n$= a^{3} + \frac{14}{5}a^{2} - \frac{7}{5}a^{3} + \frac{12a}{25}\left( 4 - 4a + a^{2} \right)\n$\n$= a^{3} + \frac{14}{5}a^{2} - \frac{7}{5}a^{3} + \frac{48}{25}a - \frac{48}{25}a^{2} + \frac{12}{25}a^{3}\n$\n$= \frac{2}{25}a^{3} + \frac{22}{25}a^{2} + \frac{48}{25}a = \frac{2}{25}\left( a^{3} + 11a^{2} + 24a \right)\n$\n$\therefore\ \frac{d\lambda}{da} = \frac{2}{25}\left( 3a^{2} + 22a + 24 \right)\n$ $\lambda$ to be greatest\\\n$\therefore\frac{d\lambda}{da} = 0 \Rightarrow 3a^{2} + 22a + 24 = 0 \Rightarrow a = - 6, - \frac{4}{3}\n$ $\frac{d^{2}\lambda}{\ da^{2}} = \frac{2}{25}(6a + 22)$, For\n$a = - 6,\frac{d^{2}\lambda}{da^{2}} < 0$\\\nSo, $\lambda$ is maximum\\\nFor $a = \frac{- 4}{3},\frac{d^{2}\lambda}{da^{2}} > 0,\lambda$ is\nminimum $\therefore\ d = \frac{2 - ( - 6)}{5} = \frac{8}{5}$