A small block slides down on a smooth inclined plane, starting from rest at time t = 0 . Let S n be the…

As we know that the displacement covered by particle in $n^{th}$second is given by,

$\because S_{n^{th}}=u+\left(\frac{2n-1}{2}\right)a$

but $u=0$

$\therefore s_n=\left(\frac{2n-1}{2}\right)a \dots \left(1\right)$

$s_{n+1}=\left[\frac{2(n+1)-1}{2}\right]a \dots \left(2\right)$

Now,
equation $\left(1\right)$ divide by equation $\left(2\right)$

$\frac{s_n}{s_{n+1}}=\frac{2n-1}{2n+1}$