Difference between revisions of "2011 AMC 12A Problems/Problem 8"

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== Solution ==
 
== Solution ==
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Let <math>A=x</math>. Then from <math>A+B+C=30</math>, we find that <math>B=25-x</math>. From <math>B+C+D=30</math>, we then get that <math>D=x</math>. Continuing this pattern, we find <math>E=25-x</math>, <math>F=5</math>, <math>G=x</math>, and finally <math>H=25-x</math>. So <math>A+H=x+25-x=25 \rightarrow \boxed{textbf{C}}</math>
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=7|num-a=9|ab=A}}
 
{{AMC12 box|year=2011|num-b=7|num-a=9|ab=A}}

Revision as of 19:10, 10 February 2011

Problem

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 $A+H$?

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

Solution

Let $A=x$. Then from $A+B+C=30$, we find that $B=25-x$. From $B+C+D=30$, we then get that $D=x$. Continuing this pattern, we find $E=25-x$, $F=5$, $G=x$, and finally $H=25-x$. So $A+H=x+25-x=25 \rightarrow \boxed{textbf{C}}$

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 12 Problems and Solutions