The sum of all natural numbers between 100 and 200, which are multiples of 3 isa.4950b.4900c.5000d.4980

To solve this problem, we first need to find the first and the last multiples of 3 between 100 and 200.

The first multiple of 3 in this range is 102 (as 100/3 = 33 remainder 1, so the next multiple is 333 + 3 = 102) and the last multiple is 198 (as 200/3 = 66 remainder 2, so the last multiple is 663 = 198).

The sum of an arithmetic series is given by the formula: sum = n/2 * (a + l) where n is the number of terms, a is the first term and l is the last term.

The number of terms can be found by the formula: n = (l - a)/3 + 1. Substituting the values we have: n = (198 - 102)/3 + 1 = 33.

Now, we can find the sum: sum = 33/2 * (102 + 198) = 4950.

So, the sum of all natural numbers between 100 and 200, which are multiples of 3 is 4950 (option a).