2011 AMC 12A Problems/Problem 8

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, because $C=5$ , $A+B+C+D+E+F+G+H=85$ .

Subtracting, we have that $A+H=25$ .

2011 AMC 12A (Problems • Answer Key • Resources)
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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