2011 AMC 10A Problems/Problem 17

Revision as of 12:58, 4 June 2019 by Ggs (talk | contribs) (Solution 2)

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

Solution 2

Given that the sum of 3 consecutive terms is 30, we have $(A+B+C)+(C+D+E)+(F+G+H)=90$ and $(B+C+D)+(E+F+G)=60$

It follows that $A+B+C+D+E+F+G+H=85$ because $C=5$.

Subtracting, we have that $A+H=25\rightarrow \boxed{\textbf{C}}$.

Solution 3

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 $H=B$, $A+H=A+B=\boxed{25 \ \mathbf{(C)}}$.

See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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