A body moves on a frictionless plane starting from rest. If is distance moved between and and is distance moved between and then the ratio is for The value of is

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Solution

Q.
A body moves on a frictionless plane starting from rest. If $S_{n}$ is distance moved between $t = n - 1$ and $t = n$ and $S_{n - 1}$ is distance moved between $t = n - 2$ and $t = n - 1,$ then the ratio $\frac{S_{n - 1}}{S_{n}}$ is $(1 - \frac{2}{x})$ for $n = 10 .$ The value of $x$ is __________ . [2024]

Ans.
(19) $S_{n} = \frac{1}{2} a {(n)}^{2} - \frac{1}{2} a {(n - 1)}^{2} = \frac{1}{2} a (2 n - 1)$

$S_{n - 1} = \frac{1}{2} a (2 \times (n - 1) - 1) = \frac{1}{2} a (2 n - 3)$

$\frac{S_{n - 1}}{S_{n}} = \frac{2 n - 3}{2 n - 1} = 1 - \frac{2}{2 n - 1} = 1 - \frac{2}{20 - 1} = 1 - \frac{2}{19}$

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