(a) Here, $s_{n} = \frac{1}{n}an^{2}$
$s_{nth}$ = distance travelled in $n$ second – distance travelled in $(n - 1)$ second
$= \left( \frac{2n - 1}{2} \right)a$
$\therefore\ \frac{s_{nth}}{s_{n}} = \frac{2n - 1}{n^{2}} = \frac{2}{n} - \frac{1}{n^{2}}$