|
|
| (7 intermediate revisions by 4 users not shown) |
| Line 1: |
Line 1: |
| − | ==Problem 17==
| + | #redirect [[2011 AMC 12A Problems/Problem 8]] |
| − | In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?
| |
| − | | |
| − | <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>.
| |
| − | | |
| − | == Solution 2 ==
| |
| − | | |
| − | We see that <math>A+B+C=30</math>, and by substituting the given <math>C=5</math>, we find that <math>A+B=25</math>. Similarly, <math>B+D=25</math> and <math>D+E=25</math>.
| |
| − | | |
| − | <cmath>\begin{align*}
| |
| − | &(A+B)-(B+D)=A-D=0\\
| |
| − | &A=D\\
| |
| − | &(B+D)-(D+E)=B-E=0\\
| |
| − | &B=E\\
| |
| − | &A, B, 5, A, B, 5, G, H
| |
| − | \end {align*}</cmath>
| |
| − | | |
| − | Similarly, <math>G=A</math> and <math>H=B</math>, giving us <math>A, B, 5, A, B, 5, A, B</math>. Since <math>H=B</math>, <math>A+H=A+B=\boxed{25 \ \mathbf{(C)}}</math>.
| |
| − | | |
| − | ===Solution 3===
| |
| − | Let <math>A=x</math>. Then from <math>A+B+C=30</math>, we find that <math>B=25-x</math>. From <math>B+C+D=30</math>, we then get that <math>D=x</math>. Continuing this pattern, we find <math>E=25-x</math>, <math>F=5</math>, <math>G=x</math>, and finally <math>H=25-x</math>. So <math>A+H=x+25-x=25 \rightarrow \boxed{\textbf{C}}</math>
| |
| − | | |
| − | | |
| − | ===Solution 4===
| |
| − | Given that the sum of 3 consecutive terms is 30, we have
| |
| − | <math>(A+B+C)+(C+D+E)+(F+G+H)=90</math> and <math>(B+C+D)+(E+F+G)=60</math>
| |
| − | | |
| − | It follows that <math>A+B+C+D+E+F+G+H=85</math> because <math>C=5</math>.
| |
| − | | |
| − | Subtracting, we have that <math>A+H=25\rightarrow \boxed{\textbf{C}}</math>.
| |
| − | | |
| − | == See Also ==
| |
| − | | |
| − | | |
| − | {{AMC10 box|year=2011|ab=A|num-b=16|num-a=18}}
| |
| − | {{MAA Notice}}
| |
