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| − | ==Problem 17==
| + | #redirect [[2011 AMC 12A Problems/Problem 8]] |
| − | In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?
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| − | <math>\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43</math>
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| − | == Solution ==
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| − | We consider the sum <math>A+B+C+D+E+F+G+H</math> and use the fact that <math>C=5</math>, and hence <math>A+B=25</math>.
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| − | <cmath>\begin{align*}
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| − | &A+B+C+D+E+F+G+H=A+(B+C+D)+(E+F+G)+H=A+30+30+H=A+H+60\\
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| − | &A+B+C+D+E+F+G+H=(A+B)+(C+D+E)+(F+G+H)=25+30+30=85
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| − | \end{align*}</cmath>
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| − | Equating the two values we get for the sum, we get the answer <math>A+H+60=85</math> <math>\Longrightarrow</math> <math>A+H=\boxed{25 \ \mathbf{(C)}}</math>.
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| − | ==Solution 2==
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| − | Given that the sum of 3 consecutive terms is 30, we have
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| − | <math>(A+B+C)+(C+D+E)+(F+G+H)=90</math> and <math>(B+C+D)+(E+F+G)=60</math>
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| − | It follows that <math>A+B+C+D+E+F+G+H=85</math> because <math>C=5</math>.
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| − | Subtracting, we have that <math>A+H=25\rightarrow \boxed{\textbf{C}}</math>.
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| − | == Solution 3 ==
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| − | We see that <math>A+B+C=30</math>, and by substituting the given <math>C=5</math>, we find that <math>A+B=25</math>. Similarly, <math>B+D=25</math> and <math>D+E=25</math>.
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| − | <cmath>\begin{align*}
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| − | &(A+B)-(B+D)=A-D=0\\
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| − | &A=D\\
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| − | &(B+D)-(D+E)=B-E=0\\
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| − | &B=E\\
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| − | &A, B, 5, A, B, 5, G, H
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| − | \end {align*}</cmath>
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| − | Similarly, <math>G=A</math> and <math>H=B</math>, giving us <math>A, B, 5, A, B, 5, A, B</math>. Since <math>H=B</math>, <math>A+H=A+B=\boxed{25 \ \mathbf{(C)}}</math>.
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| − | == Solution 4 ==
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| − | Arbitrarily select a value for B or D, and then deduce the rest using the fact that consecutive triplets add to 30. You will find that <math>A+H= \boxed{\textbf{25}}</math>.
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| − | == See Also ==
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| − | {{AMC10 box|year=2011|ab=A|num-b=16|num-a=18}}
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| − | {{MAA Notice}}
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