A small block slides down on a smooth inclined plane, starting from...

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

$∵ S{n}^{th} = u + \left(\dfrac{2n - 1}{2}\right)a$

but $u = 0$

$∴ {s}_{n} = \left(\dfrac{2n - 1}{2}\right)a \ldots \left(1\right)$

${s}_{n + 1} = \left[\dfrac{2 ( n + 1 ) - 1}{2}\right]a \ldots ( 2 )$

Now,
equation $( 1 )$ divide by equation $( 2 )$

$\dfrac{{s}_{n}}{{s}_{n + 1}} = \dfrac{2n - 1}{2n + 1}$