So we have: a1 + a2 + ... + a100 = 15,000.
We can represent each term as:
a1 = a1
a2 = a1 + q
a3 = a2 + q = a1 + 2q
.
.
an = a1 + (n-1)*q

Using this, the first equation is equal to:

(1) 100a1 + (99100/2)*q = 15,000

(We are adding q's as: q + 2q + 3q + .. + 99q. That is a sum of integers, and it's equal to n(n+1)/2, or 99*100/2)

We are also given the following: a3 + a6 + a9 + .. + a99 = 5016. So it's every third (there are 33 of them).
Using the representation from above, this equation is equal to:

(2) 33a1 + 1650q = 5016.

This time, we are adding q's as: 2q + 5q + .. + 98q. This would be the following [; \sum{i=1}^{33} 3i-1 ;] which you can simplify:
[; \sum{i=1}^{33} 3i-1 = 3\sum{i=1}^{33} i - \sum{i=1}^{33} 1 = 3 * (3334)/2 - 33 = 1650;]

(The first sum, is the sum of integers, so we can calculate it using the n(n+1)/2 formula, and the second one is sum of 33 1's, so we have 3 * 3334/2 - 33 = 1650)

So now we have two very simple equations:

a1 + 49.5q = 150 (if we divide the first one with 100)
33a1 + 1650*q = 5016

When we solve the system, the solution is a1 = -48 and q = 4
Now, we can use the formula for the sum of an arithmetic sequence: Sn = n(a1 + an)/2 (where n=100, a1=-48 and an = a1 + 99q = 348) to check if we got the right answer. And we did!

Now, I haven't done this type of things in ages, so I may have over-complicated it or it might not be correct, but it seems okay to me.