This one is a simple arithmetic progression problem.
first time = a = 100.
last term = l = 300
difference = d = 2 (as alternate terms are there, 100, 102, 104)
now, l = a + (n-1) * d
=> 300 = 100 + (n-1) * 2
=> n = 101
now sum = S = (n/2) * [2a + (n-1) * d ]
=> S = (101/2) * [2*100 + (101-1) *2 ]
=> S = (101/2)*[400] = 101 * 200 = 20200
=> S = 20200
