Explanation

(d)
Since, in a binomial expansion of $(a - b)^{n},\ n \geq 5$, then sum of 5th and 6th terms is equal to zero.
$\ \therefore\ \ \ \ \ \ \ \ \ ^{n}C_{4}a^{n - 4}({- b)}^{4} + \ ^{n}C_{5}a^{n - 5}({- b)}^{5} = 0$
$\Longrightarrow \frac{n!}{(n - 4)!4!}a^{n - 4}b^{4} - \frac{n!}{(n - 5)!5!}a^{n - 5}b^{5} = 0$
$\Longrightarrow \frac{n!}{(n - 5)!4!}a^{n - 5} \bullet b^{4}\left( \frac{a}{n - 4} - \frac{b}{5} \right) = 0$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \frac{a}{b} = \frac{n - 4}{5}$