Fly and Two Trains Riddle 
Two trains travel on the same track towards each other, each going at a speed of 50 kph. They start out 100km apart. A fly starts at the front of one train and flies at 75 kph to the front of the other; when it gets there, it turns around and flies back towards the first. It continues flying back and forth till the two trains meet and it gets squashed.
How far did the fly travel before it got squashed?

The question is relatively simple - the trains take an hour to collide, and in an hour the fly would travel 75 km. However, the question got me wondering: Lets say you are given a time $t$ in hours(where $t< 1$) How many "rounds" would the fly make in $t$ time? A round  is defined as moving from one train to another and back. How would I model the equation?
 A: The time taken to do the $n^{th}$ round is $\dfrac{a(n)}{(75+50)}$ where $a(n)$ is the distance between the two trains at the beginning of the $n^{th}$ round.
Note that $$a(n+1) = a(n) - (50+50) \times \dfrac{a(n)}{125} = \dfrac{a(n)}{5} = \dfrac{a(1)}{5^n}.$$
The distance travelled by the fly in the $n^{th}$ round is $75 \times \dfrac{a(n)}{125} = \dfrac{3}{5} a(n)$.
Hence, the total distance travelled by the fly is $$\sum_{n=1}^{\infty} \dfrac{3}{5} a(n) = a(1) \dfrac{3}{5} \left( 1 + \dfrac15 + \dfrac1{5^2} + \cdots \right) = a(1) \dfrac{3}{5} \times \dfrac{5}{4} = \dfrac34 a(1)$$ which is nothing but $$\text{Speed of the fly} \times \underbrace{\dfrac{a(1)}{\text{Relative speed between the trains}}}_{\text{Time taken by the trains to collide}}$$
which makes sense.
If you want the total number of rounds the fly makes before the train collides, this amounts to the number of rounds till when $a(n) = 0$. However note that only as $n \rightarrow \infty$, $a(n) \rightarrow 0$. Hence, the fly will make $\infty$-rounds before the trains collide!
The time taken by the fly to make the first round is $$\dfrac{a(1)}{125}.$$ The time taken by the fly to make the $n^{th}$ round is $$\dfrac{a(n)}{125}.$$ Hence, the time taken till the $N^{th}$ round is $$\sum_{n=1}^{N} \dfrac{a(n)}{125} = \sum_{n=1}^{N} \dfrac{ \dfrac{a(1)}{5^{n-1}}}{125}$$
Hence, solve for $N$ in the equation $$\sum_{n=1}^{N} \dfrac{ \dfrac{a(1)}{5^{n-1}}}{125} = t$$
Setting $a(1) = 100$, we get that
$$\sum_{n=0}^{N-1} \left(\dfrac15 \right)^n = \dfrac{5t}{4}$$
$$\dfrac{\left(\dfrac15 \right)^N - 1}{\left(\dfrac15 \right) - 1} = \dfrac{5t}{4}$$
$$1 - \left(\dfrac15 \right)^N = t$$ $$\left(\dfrac15 \right)^N = 1-t$$ Hence the number of rounds, $N$, made by the fly as a function of time $t$ is $$N = \dfrac{\log(1-t)}{\log(1/5)}$$
