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(C) The distance travelled by an object in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.Since the block starts from re

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$\frac{2n - 1}{2n + 1}$

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(C) The distance travelled by an object in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.Since the block starts from rest,the initial velocity $u = 0$.Therefore,the distance travelled in the $n^{th}$ second is $S_n = \frac{a}{2}(2n - 1)$.Similarly,the distance travelled in the $(n+1)^{th}$ second is $S_{n+1} = \frac{a}{2}(2(n+1) - 1) = \frac{a}{2}(2n + 2 - 1) = \frac{a}{2}(2n + 1)$.Taking the ratio of these two distances:$\frac{S_n}{S_{n+1}} = \frac{\frac{a}{2}(2n - 1)}{\frac{a}{2}(2n + 1)} = \frac{2n - 1}{2n + 1}$.

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

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