A particle starts sliding down a frictionless include plane. If $S_{n}$ is the distance travelled by it from time $t = n - 1 s e c o n d$ $t = n s e c o n d$ , the ratio $S_{n} / S_{n + 1}$ is

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{2 n - 1}{2 n + 1}$
Distance travelled in n th second = $u + \frac{a}{2} (2 n - 1)$
As body starts sliding so $u = 0$

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

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

Suggest Corrections

Similar questions

S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}

t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}},

\frac{t_{n}}{S_{n}}

1 - 2 n + \frac{2 n (2 n - 1)}{2!} - \frac{2 n (2 n - 1) (2 n - 1)}{3!} + . . . + (- 1)^{n - 1} \frac{2 n (n - 1) . . . (n + 2)}{(n - 1)}

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

a = \infty \sum n = 1 \frac{2 n}{(2 n - 1)!}, b = \infty \sum n = 1 \frac{2 n}{(2 n + 1)!}

a b

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 1}

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 2}

S_{n} = \frac{1}{2 n} + \frac{1}{\sqrt{4 n^{2} - 1}} + \frac{1}{\sqrt{4 n^{2} - 4}} + . . . . . . . . + \frac{1}{\sqrt{3 n^{2} + 2 n - 1}}, n \in N

{lim}_{n \to \infty} S_{n} = α

Join BYJU'S Learning Program