100 people are standing in a row such that the distance between the person standing at the $$n^{th}$$ place and the person standing at the $$(n + 1)^{th}$$ place is exactly (n + 1) metres. What is the distance between the person standing at the $$1^{st}$$ place and the person standing at the $$50^{th}$$ place in the row (in metres)?

The distance between the 1st position and 2nd position would be 2 meters. 
The distance between the 2nd position and 3rd position would be 3 meters.
The distance between the 3rd position and the 4th position would be 4 meters. 
we can see a pattern emerging here, 
.....
The distance between the 49th position and the 50th position would be 50 meters. 

So the total distance between the 1st position and the 50th position would be 2+3+4+......+50
We can easily calculate this as the sum of an AP and it turns out to be $$\frac{49}{2}\left[2\left(2\right)+\left(48\times\ 1\right)\right]$$
Which is equal to 1274

Hence, Option A is the correct answer.