Explanation

(d) $\because v = 0 + na \Rightarrow a = v/n$
Now, distance travelled in $n\ sec$.$\Rightarrow S_{n} = \frac{1}{2}an^{2}$ and
distance travelled in $(n - 2)sec \Rightarrow S_{n -^{2}} = \frac{1}{2}a(n - 2)^{2}$
$\therefore Distance\ travelled\ in\ last\ 2\ seconds,$
$= S_{n} - S_{n - 2\ }$=$\frac{1}{2}an^{2} - \frac{1}{2}a(n - 2)^{2}$
$\frac{a}{2}\left\lbrack n^{2} - (n - 2)^{2} \right\rbrack = \frac{a}{2}\left\lbrack n + (n - 2) \right\rbrack\left\lbrack n - (n - 2) \right\rbrack$
$= a(2n - 2) = \frac{v}{n}(2n - 2) = \frac{2v(n - 1)}{n}$