# 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 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

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 (the tedious one)

From the given information, we can deduce the following equations: $A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G =30$, and $F+G+H=30$.

We can then cleverly manipulate the equations above by adding and subtracting them to be left with the answer. $(A+B)-(B+D)=25-25 \implies (A-D)=0$ $(A-D)+(D+E)=0+25 \implies (A+E)=25$ $(A+E)-(E+F+G)=25-30 \implies (A-F-G)=-5$ (Notice how we don't use $D+E+F=30$) $(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25$

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

~JinhoK

## Solution 4 (the cheap one)

Since all of the answer choices are constants, it shouldn't matter what we pick $A$ and $B$ to be, so let $A = 20$ and $B = 5$. Then $D = 30 - B -C = 20$, $E = 30 - D - C = 5$, $F = 30 - D - E =5$, and so on until we get $H = 5$. Thus $A + H = \boxed{\mathbf{(C)}25}$

## Note

Something useful to shorten a lot of the solutions above is to notice $$5 + D + E = D + E + F$$ so F = 5

## Podcast Solution

https://www.buzzsprout.com/56982/episodes/383730 (Episode starts with a solution to this question) ~Wes Carroll

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 