Explanation

(a)
In the expansion of $(1 + x)^{n}$,it is given that $\ ^{n}C_{1},\ ^{n}C_{2},\ ^{n}C_{3}$ are in AP
$\Rightarrow 2\ ^{n}C_{2} = \ ^{n}C_{1} + \ ^{n}C_{3}$
$\Rightarrow 2 \bullet \frac{n(n - 1)}{1 \bullet 2} = \frac{n}{1} + \frac{n(n - 1)(n - 2)}{1 \bullet 2 \bullet 3}$
$\Rightarrow \ 6(n - 1) = 6 + (n - 2)(n - 1)$
$\Rightarrow \ 6n - 6 = 6 + n^{2} - 3n + 2$
$\Rightarrow \ n^{2} - 9n + 14 = 0$
$\Rightarrow \ (n - 2)(n - 7) = 0$
$\Rightarrow \ n = 2,\ 7$
But $n = 2$ is not acceptable because, when $n = 2$, there are only three terms in the expansion of $(1 + x)^{2}$
$\therefore\ n = 7$