Q. A small block slides without friction down an inclined plane starting from rest. Let sn 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)$
$\displaystyle \frac{S_n}{S_{n+1}}= \frac{2n-1}{2n+1}$