2011 AMC 10A Problem 17

We know only one term, $C = 5$. The trick is to compare two groups of three that overlap.

1. Compare two neighbouring groups. $A + B + C = 30$ and $B + C + D = 30$. Both groups contain $B$ and $C$. Since the totals are equal, the leftover terms must be equal too: $A = D$.
2. Keep sliding. Comparing $B + C + D$ with $C + D + E$ gives $B = E$. Comparing $C + D + E$ with $D + E + F$ gives $C = F$. And so on. Each term equals the term three places after it.
3. So the sequence repeats every three terms:
$$A,\ B,\ 5,\ A,\ B,\ 5,\ A,\ B.$$
In particular, the last term $H$ equals $B$.
4. Finish. $A + H = A + B$. From the first group, $A + B + 5 = 30$, so $A + B = 25$.

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

When every block of three neighbouring terms has the same sum, sliding the block one step drops one term and adds another. For the sum to stay the same, the dropped term and the added term must be equal. So the sequence repeats with a cycle of three. Once you see the repeat, you can move any term back into the first block, where you know the total.

Try a concrete example. Pick $A = 10$; then $B = 30 - 10 - 5 = 15$. Next $D = 30 - B - C = 30 - 15 - 5 = 10$, $E = 30 - C - D = 15$, $F = 5$, $G = 10$, $H = 15$. So $A + H = 10 + 15 = 25$. Trying a different starting value such as $A = 0$ gives $B = 25$, $H = 25$, and again $A + H = 25$, which suggests (and the main solution proves) that the answer does not depend on the choice.

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 $\boxed{\textbf{(C)}\ 25}$.