2002 AMC 10B Problem 19

Let $d = a_2 - a_1$ be the common difference. Moving $100$ places along an arithmetic sequence adds $100d$, so $a_{k+100} = a_k + 100d$ for every $k$.

Pair the $k$-th term of the second block with the $k$-th term of the first block. Each of the $100$ pairs differs by $100d$, so subtracting the two given sums,
$$\sum_{k=1}^{100} \left(a_{k+100} - a_k\right) = 100 \cdot 100d = 10000d.$$
The left side equals $200 - 100 = 100$, so $10000d = 100$ and $d = \frac{1}{100} = 0.01$.

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

In an arithmetic sequence the difference between terms depends only on how far apart they are, so two blocks of equal length that are offset by $m$ positions have sums differing by (block length) $\times\ m d$. Subtracting the block sums eliminates $a_1$ automatically; there is no need for the full sum formula.

Sum formula: $100a_1 + (0 + 1 + \cdots + 99)d = 100a_1 + 4950d = 100$, and for the second block $100a_{101} + 4950d = 200$ with $a_{101} = a_1 + 100d$. Subtracting gives $100\cdot 100d = 100$, the same equation.

• Stopping at 100d = 1 and answering 1 (choice E), which is the gap between terms one hundred apart, not between consecutive terms.
• Setting up $\frac{n}{2}(2a_1 + (n-1)d)$ with $n = 200$ and $n = 100$ and making an arithmetic slip in the difference of the $d$ coefficients.