Explanation

As we know, distance traversed in $n^{th} \text{second}$ $S_{n} = u + \frac{1}{2}a(2n - 1)$ Here, $u = 0$, $a = g$ $\therefore S_{n} = \frac{1}{2}g(2n - 1)$ Distance traversed in $1^{st} \text{second}$ i.e., $n = 1$ $S_{1} = \frac{1}{2}g(2 \times 1 - 1) = \frac{1}{2}g$ Distance traversed in $2^{nd} \text{second}$ i.e., $n = 2$ $S_{2} = \frac{1}{2}g(2 \times 2 - 1) = 3/2 g$ Distance traversed in $3^{rd} \text{second}$ i.e., $n = 3$ $S_{3} = \frac{1}{2}g(2 \times 3 - 1) = 5/2 g$ $\therefore S_{1}:S_{2}:S_{3} = \frac{1}{2}g:3/2 g:5/2 g = 1:3:5$