List them,

201,204,207,...,498

This is arithmetic sequence we commond difference 3.

Thus,

a_n=3n+b

Where, n is the "n-th" term.

Thus,

a_1=3(1)+b

But,

a_1=201

Thus,

201=3+b

And hence,

b=198

In general this sequence can be modelled as,

a_n=3n+198

Let us find what "n" is in the last term,

498=3n+198

300=3n

n=100

Thus, there are 100 terms.

Hence,

a_n=3n+198 where "n" ranges from 1 to 100.

We need to find the sum,

a_1+a_2+...+a_100

That is,

3(1)+198+3(2)+198+3(3)+198+...+3(100)+198

Rewrite this as,

3(1+2+...+100)+(198+198+...+198)

Where 198 appears 100 terms.

Thus,

3(1+2+...+100)+100*198

Now use the formula,

1+2+...+n=n(n+1)/2

Hence,

3(5050)+100*198

Is the answer.