Solution:
The distance in the \(n^{\text{th}}\) second is \(S_n = u + \frac{a}{2}(2n-1)\). Since \(u=0\), \(S_n \propto 2n-1\). Thus, \(\frac{S_n}{S_{n+1}} = \frac{2n-1}{2(n+1)-1} = \frac{2n-1}{2n+1}\).
A small block slides down on a smooth inclined plane, starting from rest at time \(t = 0\). Let \(S_n\) be the distance travelled by the block in the interval \(t = n - 1\) to \(t = n\). Then, the ratio \(\frac{S_n}{S_{n+1}}\) is
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