2011 AMC 10A Problems/Problem 17

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 $A+H$ ?

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

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 $A+H+60=85$ $\Longrightarrow$ $A+H=\boxed{25 \ \mathbf{(C)}}$ .

We see that $A+B+C=30$ , and by substituting the given $C=5$ , we find that $A+B=25$ . Similarly, $B+D=25$ and $D+E=25$ .

$\begin{align*} &(A+B)-(B+D)=A-D=0\\ &A=D\\ &(B+D)-(D+E)=B-E=0\\ &B=E\\ &A, B, 5, A, B, 5, G, H \end {align*}$

Similarly, $G=A$ and $H=B$ , giving us $A, B, 5, A, B, 5, A, B$ . Since $H=B$ , $A+H=A+B=\boxed{25 \ \mathbf{(C)}}$ .

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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