Q. A small block slides without friction down an inclined plane starting from rest. Let *s*_{n} be the distance travelled from *t* = *n*-1 to *t* = *n*. Then $ \displaystyle \frac{S_n}{S_{n+1}}$ is:

(a) $ \displaystyle \frac{2n-1}{2n}$

(b) $ \displaystyle \frac{2n+1}{2n-1}$

(c) $ \displaystyle \frac{2n-1}{2n+1}$

(d) $ \displaystyle \frac{2n}{2n+1}$

Ans: (c)

Sol:

$\displaystyle S_n = 0+\frac{a}{2}(2n-1) $

$\displaystyle S_{n+1} = 0+ \frac{a}{2}(2(n+1)-1) $

$ \displaystyle S_{n+1} = \frac{a}{2}(2n+1)$

$\displaystyle \frac{S_n}{S_{n+1}}= \frac{2n-1}{2n+1}$