Difference between revisions of "2011 AMC 10A Problems/Problem 17"

Revision as of 09:47, 8 May 2011

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 $A+H$ ?

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

We consider the sum $A+B+C+D+E+F+G+H$ and use the fact that $C=5$ , and hence $A+B=25$ .

$\begin{align*} &A+B+C+D+E+F+G+H=A+(B+C+D)+(E+F+G)+H=A+30+30+H=A+H+60\\ &A+B+C+D+E+F+G+H=(A+B)+(C+D+E)+(F+G+H)=25+30+30=85 \end{align*}$

Equating the two values we get for the sum, we get the answer $A+H+60=85$ $\Longrightarrow$ $A+H=\boxed{25 \ \mathbf{(C)}}$ .

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 18
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 10 Problems and Solutions

@@ Line 14: / Line 14: @@
 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>.
+== See Also ==
+{{AMC10 box|year=2011|ab=A|num-b=17|num-a=18}}