Explanation

(d) As $v = 0 + na$
$\Rightarrow \ \ \ \ \ \ \ \ \ a = \frac{v}{n}$
Now, distance travelled in $n$ sec
$\Rightarrow \ \ \ \ \ \ \ S_{n} = \frac{1}{2}an^{2}$
Distance travelled in $(n - 2)$ sec
$\Rightarrow \ \ \ {\ S}_{n - 2} = \frac{1}{2}\ a(n - 2)^{2}$
Distance travelled in last 2 s,
$S_{n} - S_{n - 2} = \frac{1}{2}\ an^{2} - \frac{1}{2}a\ (n - 2)^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{a}{2}\left\lbrack n^{2} - (n - 2)^{2} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{a}{2}\left\lbrack n + (n - 2) \right\rbrack\left\lbrack n - (n - 2) \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = a(2n - 2) = \frac{v}{n}(2n - 2) = \frac{2v(n - 1)}{n}$