Explanation

(b)
Here, $T_{4} = \ ^{n}C_{3}(a)^{n - 3}( - 2b)^{3}$
and $T_{5} = \ ^{n}C_{4}(a)^{n - 4}( - 2b)^{4}$
Given, $T_{4} + T_{5} = 0$
$\Rightarrow \ \ ^{n}C_{3}{(a)}^{n - 3}{( - 2b)}^{3} + \ ^{n}C_{4}{(a)}^{n - 4}{( - 2b)}^{4} = 0$
$\Rightarrow \ {(a)}^{n - 4}{( - 2b)}^{3}\left\lbrack a\ ^{n}C_{3} + \ ^{n}C_{4}( - 2b) \right\rbrack = 0$
$\Rightarrow \ \frac{a}{b} = \frac{2\ ^{n}C_{4}}{\ ^{n}C_{3}}$
$= \frac{2.n(n - 1)(n - 2)(n - 3)}{4.3.2.1} \times \frac{3.2.1}{n(n - 1)(n - 2)}$
$= \frac{n - 3}{2}$