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2011 AMC 10A Problems/Problem 17

Revision as of 17:55, 14 February 2011 by Flyingpenguin (talk | contribs)

Problem 17

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 $A+H$?

$\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43$

Solution

We consider the sum $A+B+C+D+E+F+G+H$ and use the fact that $C=5$, and hence $A+B=25$.

\begin{align*} &A+B+C+D+E+F+G+H=A+(B+C+D)+(E+F+G)+H=A+30+30+H=A+H+60\\ &A+B+C+D+E+F+G+H=(A+B)+(C+D+E)+(F+G+H)=25+30+30=85 \end{align*}

Equating the two values we get for the sum, we get the answer $A+H+60=85$ $\Longrightarrow$ $A+H=\boxed{25 \ \mathbf{(C)}}$.

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