2011 AMC 10A Problem 17

P16-20~3 minPrint

Sequences & Series Algebraic Manipulation Set up the equation/formula and compute; no special trick needed Compute small cases, spot the pattern, generalize

In the eight-term sequence $A,B,C,D,E,F,G,H$ , the value of $C$ is 5 and the sum of any three consecutive terms is 30. What is $A+H$ ?

Show solution without answering

Key insight

Two overlapping groups of three share two terms, so each term equals the term three places later. So H equals B, and A + B = 30 − 5.

Solution

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

Why this works

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.

Alternative approach

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)