Explanation

(d)
Given expansion is $\left( x - \frac{1}{2x} \right)^{n}$
$\therefore\ T_{3} = \ ^{n}C_{2}{(x)}^{n - 2}\left( - \frac{1}{2x} \right)^{2}$
and $T_{4} = \ ^{n}C_{3}{(x)}^{n - 3}\left( - \frac{1}{2x} \right)^{3}$
But according to the given condition,
$\frac{T_{3}}{T_{4}} = - \frac{n(n - 1) \times 3 \times 2 \times 1 \times 8}{n(n - 1)(n - 2) \times 2 \times 1 \times 4 \times} = \frac{1}{2}$ (given)
$\Rightarrow \ - n + 2 = 12 \Rightarrow n = - 10$