2011 AMC 10A Problems/Problem 17: Difference between revisions

Revision as of 17:55, 14 February 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 $A+H+60=85$ $\Longrightarrow$ $A+H=\boxed{25 \ \mathbf{(C)}}$ .

@@ Line 3: / Line 3: @@
 <math>\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43</math>
+== Solution ==
+We consider the sum <math>A+B+C+D+E+F+G+H</math> and use the fact that <math>C=5</math>, and hence <math>A+B=25</math>.
+<cmath>\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*}</cmath>
+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>.