First way to solve the question is by grouping:

2 + 4 + 6 + 8 + ....+ 198 + 200 =

= (2 + 200) + (4 + 1998) + ... + (100 + 102) = 50*202 = 10,100

Second way to calculate this sum is to consider the arithmetic progression:

a1 = 2 (since even, so it will start with 2)

d = 2 (the common difference)

an = an-1 + 2

an = 2n

The sum of the first n terms is:

Sn = (n/2)(a1 + an)

In our case we need the sum of the first 100 terms, that is, when n = 100:

S100 = (n/2)[a1 + a100]

S100 = (100/2)[2 + 200]

S100 = 10,100

In both cases we get the same answer: the sum of the first 100 positive even numbers is 10,100.