2011 AMC 12A Problems/Problem 8

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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

Solution 1

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 $A+H=x+25-x=25 \rightarrow \boxed{\textbf{C}}$

Solution 2

A faster technique is to assume that the problem can be solved, and thus $A+H$ is an invariant. Since $A + B + 5 = 30$, assign any value to $A$. $10$ is a simple value to plug in, which gives a value of $15$ for B. The 8-term sequence is thus $10, 15, 5, 10, 15, 5, 10, 15$. The sum of the first and the last terms is $25\rightarrow \boxed{\textbf{C}}$

Note that this alternate solution is not a proof. If the sum of $A+G$ had been asked for, this technique would have given $20$ as an answer, when the true answer would have been "cannot be determined".

Solution 3

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 $A+H=25\rightarrow \boxed{\textbf{C}}$.

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

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