Explanation

(b)
Let $a = \ ^{n}C_{r - 1},\ b = \ ^{n}C_{r},c = \ ^{n}C_{r + 1}$
and $d = \ ^{n}C_{r + 2}$
$\therefore\ a + b = \ ^{n + 1}C_{r},\ b + c = \ ^{n + 1}C_{r + 1},\ c + d = \ ^{n + 1}C_{r + 2}$
$\Rightarrow \ \frac{a + b}{a} = \frac{\ ^{n + 1}C_{r}}{\ ^{n}C_{r - 1}} = \frac{n + 1}{r}\ \Rightarrow \ \frac{a}{a + b} = \frac{r}{n + 1}$
and $\ \frac{b + c}{b} = \frac{\ ^{n + 1}C_{r + 1}}{\ ^{n}C_{r}} = \frac{n + 1}{r + 1}\ \Rightarrow \ \frac{b}{b + c} = \frac{r + 1}{n + 1}$
and $\frac{c + d}{c} = \frac{\ ^{n + 1}C_{r + 2}}{\ ^{n}C_{r + 1}} = \frac{n + 1}{r + 2}\ \Rightarrow \ \frac{c}{c + d} = \frac{r + 2}{n + 1}$
$\therefore\ \frac{a}{a + b},\ \frac{b}{b + c}$, $\frac{c}{c + d}$ are in AP
$\because AM > GM$
$\Rightarrow \ \frac{b}{b + c} > \sqrt{\frac{ac}{(a + b)(c + d)}}$
or $\left\{ \left( \frac{b}{b + c} \right)^{2} - \frac{ac}{(a + b)(c + d)} \right\} > 0$