2002 AMC 10B Problem 19

Let $d$ be the common difference, so $d = a_2 - a_1$ is what we want. Each step along the sequence adds $d$.

1. Pair each term in the second group with the term exactly $100$ places before it:
$$a_{101} \text{ with } a_1, \quad a_{102} \text{ with } a_2, \quad \ldots, \quad a_{200} \text{ with } a_{100}.$$
2. Going $100$ places adds $d$ a hundred times, so in every pair the later term is $100d$ bigger. For example, $a_{101} = a_1 + 100d$.
3. There are $100$ pairs, so the second sum is bigger than the first by
$$100 \times 100d = 10000d.$$
4. The problem says the sums are $200$ and $100$, which differ by $100$. So
$$10000d = 100, \qquad d = \frac{100}{10000} = 0.01.$$

The answer is $\boxed{\textbf{(C)}\ 0.01}$.

In an arithmetic sequence, the gap between two terms depends only on how many steps apart they are. Two blocks of the same length, shifted by the same distance, therefore differ term by term by the same amount, and subtracting the block sums makes the first term disappear. You never needed the formula for the sum of an arithmetic sequence.

Try a small version to see the pattern. With $1, 2, 3, 4$ (so $d = 1$), the first two terms add to $3$ and the next two add to $7$: a difference of $4 = 2 \times 2 \times 1$, that is, (block length) times (shift) times $d$. In our problem that is $100 \times 100 \times d = 100$, so $d = 0.01$.