Find how many 3-digit numbers leave a remainder of 2 when divided by 5, and find their sum.

Step 1. The AP. The smallest is $102$ and the largest $997$, with $d = 5$.

Step 2. How many. $997 = 102 + (n - 1) \times 5$ gives $n - 1 = 179$, so $n = 180$.

Step 3. The sum. $S = \dfrac{180}{2}(102 + 997) = 90 \times 1099 = 98910$.