A small block slides down on a smooth inclined plane, starting from the rest at time t = 0.  Let${\mathit{\text{S}}}_{\text{n}}$  be the distance travelled by the block in the interval t = n – 1 to t = n.  Then, the ratio  $\frac{{\mathit{\text{S}}}_{\text{n}}}{{\mathit{\text{S}}}_{\text{n +1}}}$ is 

Let $\theta$  be the inclination
By equation of uniformly accelerated motion, distance travelled by object during nth second is given by 
${\text{S}}_{\text{n}}=\text{u}+\frac{\text{a}}{\text{2}}\left(\text{2n}-\text{1}\right)$
$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}S}}_{\text{n}}=0+\frac{\text{g\hspace{0.17em}}\sin \theta }{2}\left(2\text{\hspace{0.17em}n}-1\right)$
${\text{S}}_{\text{n}}=\frac{\text{g\hspace{0.17em}}\sin \theta }{2}\left(2\text{\hspace{0.17em}n}-1\right)\dots . \left(i\right)$
Distance travelled during $\left(\text{n}+1\right)\text{\hspace{0.17em}th}$  second
${\text{S}}_{\text{n}+1}=0+\frac{\text{g\hspace{0.17em}}\sin \theta }{2}\left[2\left(\text{n}+1\right)-1\right]$
$=\frac{\text{g\hspace{0.17em}}\sin \theta }{2}\left(2\text{\hspace{0.17em}n}+1\right)\dots . \left(ii\right)$
Using Equations (i) and (ii), we get
$\frac{{\text{S}}_{\text{n}}}{{\text{S}}_{\text{n}+\text{1}}}=\frac{2\text{\hspace{0.17em}n}-1}{2\text{\hspace{0.17em}n}+1}$