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 ( S n / S n +1) is :

Question

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 ( S n / S n +1) is : (A) (2 n-1/2 n) (B) (2 n-1/2 n+1) (C) (2 n+1/2 n-1) (D) (2 n/2n -1)

Tardigrade Admin · Accepted Answer

S n =0+(1/2) a (2 n -1) S n +1=0+(1/2) a (2( n +1)-1) =(1/2) a (2 n +1) ( S n / S n +1)=((2 n -1)/(2 n +1))

Q. 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 :

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

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

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

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

$S_{n} = 0 + \frac{1}{2} a (2 n - 1)$
$S_{n + 1} = 0 + \frac{1}{2} a (2 (n + 1) - 1)$
$= \frac{1}{2} a (2 n + 1)$
$\frac{S _{n}}{S _{n + 1}} = \frac{( 2 n - 1 )}{( 2 n + 1 )}$

Q. 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 Sn+1​Sn​​ is :

2n2n−1​

2n+12n−1​

2n−12n+1​

2n−12n​