2011 AMC 10A Problem 17

Compare two overlapping triples: $A + B + C = 30$ and $B + C + D = 30$. Subtracting gives $A = D$. The same argument on every pair of consecutive triples shows that each term equals the term three places later:
$$A = D = G,\qquad B = E = H,\qquad C = F = 5.$$

So $H = B$, and the desired sum is
$$A + H = A + B = 30 - C = 30 - 5 = 25.$$

The answer is $\boxed{\textbf{(C)}\ 25}$.

A constant sum over a sliding window of length $k$ forces the sequence to be periodic with period $k$: subtracting adjacent window sums cancels everything but the entering and leaving terms. Once the period is known, any far-apart pair of terms can be pulled back into a single window, where the given constraint applies.

Write the sequence in terms of $A$ and $B$: $A, B, 5, A, B, 5, A, B$. Every triple of consecutive terms is a rearrangement of $\{A, B, 5\}$ and sums to $30$, so $A + B = 25$, and the last term is $H = B$.

• Trying to determine A and H individually; they are not determined, only their sum is, and A + B = 25 comes straight from A + B + C = 30.
• Assuming the sequence is arithmetic or that all terms equal $10$, which contradicts $C = 5$.