Explanation

(b)
Here, $T_{5} + T_{6} = 0$
$\Longrightarrow \ ^{n}C_{4}a^{n - 4}({- 2b)}^{4} + \ ^{n}C_{5}a^{n - 5}({- 2b)}^{5} = 0$
$\Longrightarrow 16 \bullet \ ^{n}C_{4}a^{n - 4}b^{4} = 32\ ^{n}C_{5}a^{n - 5}b^{5}$
$\Longrightarrow \frac{\ ^{n}C_{5}}{\ ^{n}C_{4}} \bullet \frac{a^{n - 5}b^{5}}{a^{n - 4}b^{4}} = \frac{1}{2}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{b}{a} = \frac{1}{2} \bullet \frac{\ ^{n}C_{4}}{\ ^{n}C_{5}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{a}{b} = \frac{2 \bullet \frac{n!}{5!(n - 5)!}}{\frac{n!}{4!(n - 4)!}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{4!(n - 4)!}{5!(n - 5)!} \times 2 = \frac{2(n - 4)}{5}$