A particle is moving along a straight line with constant acceleration. If the distance travelled by the body in $$n^{\text{th}}$$ and $$(n + 1)^{\text{th}}$$ seconds is $$100\text{ m}$$, find it's velocity (in $$\text{m/s}$$) at the end of $$n^{\text{th}}$$ second.

The distance in the $$n^{\text{th}}$$ second is $$u + a\left(n - \frac{1}{2}\right)$$ and in the $$(n + 1)^{\text{th}}$$ second is $$u + a\left(n + \frac{1}{2}\right)$$. Their sum is $$2u + 2an = 100$$, so $$u + an = 50$$. This is exactly the velocity at the end of the $$n^{\text{th}}$$ second, $$50\text{ m/s}$$.