Explanation

(b)
Coefficient of $p$th, $(p + 1)$th and $(p + 2)$th terms in the expansion ${(1 + x)}^{n}$ are $\ ^{n}C_{p - 1},\ \ ^{n}C_{p},\ \ ^{n}C_{p + 1}$ respectively
Since, these are in AP
$\therefore\ 2\ ^{n}C_{p} = \ ^{n}C_{p - 1} + \ ^{n}C_{p + 1}$
$\Rightarrow \ 2\frac{n!}{(n - p)!p!} = \frac{n!}{(n - p + 1)!(p - 1)!} + \frac{n!}{(n - p - 1)!(p + 1)!}$
$\Rightarrow \ \frac{2}{(n - p)!p!} = \frac{p}{(n - p + 1)(n - p)!p!} + \frac{n - p}{(n - p)!(p + 1)p!}$
$\Rightarrow \ \frac{2}{1} = \frac{p}{(n - p + 1)} + \frac{n - p}{p + 1}$
${\Rightarrow \ n}^{2} - n(4p + 1) + 4p^{2} - 2 = 0$