Question

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 :

Answer 1

Option : (2). \(\frac{2n-1}{2n+1}\)

S_n = Distance in n^th sec. 

i.e. t = n – 1 to t = n

S_{n + 1} = Distance in (n + 1)^th sec.

i.e. t = n to t = n + 1

So as we know,

Answer 2

Correct option is (2) \(\frac{2n -1}{2n+1}\)

For distance S_n​ travelled from (n−1) to n

S_n ​= 0[n−n−1] + \(\frac12\) ​a[n² − (n−1)²]

S_n ​= \(\frac12\)​ a[n² − (n² + 1 − 2n)]

S_n​ = \(\frac a2\)(2n − 1)

Again, S_n+1 ​= \(\frac12\) ​a[(n+1)² − n²]

= \(\frac a2\)(2n + 1)

\(\frac{S_n}{S_{n+1}} = \frac{\frac a2(2n -1)}{\frac a2(2n+1)}\)

\(= \frac{2n -1}{2n+1}\)