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A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn 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:
Option: 1
$\frac{2n-1}{2n}$

Option: 2
$\frac{2_{n}-1}{2n+1}$

Option: 3
$\frac{2_{n}+1}{2n-1}$

Option: 4
$\frac{2_{n}}{2n-1}$

Answers (1)

From $t=0$ to $t=n-1$

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

From $t=0\: to \: t=n$

$S_{2}=0+\frac{1}{2}a\left ( n \right )^{2}$

From $t=0\: to \: t=n+1$

$S_{3}=0+\frac{1}{2}a\left ( n+1 \right )^{2}$

$S_{n}=S_{2}-S_{1}$

$S_{n}=\frac{1}{2}a\left [ n^{2}-\left ( n^{2}-2n+1 \right ) \right ]$

$S_{n+1}=S_{3}-S_{2}=\frac{1}{2}a\left [ n^{2}+2n+1-n^{2} \right ]$

$\frac{S_{n}}{S_{n}+1}=\frac{\left ( 2n-1 \right )}{\left ( 2n+1 \right )}$

Posted by

Ramraj Saini

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