Difference between revisions of "2011 AMC 10A Problems/Problem 17"
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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>. | 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== | |
Given that the sum of 3 consecutive terms is 30, we have | 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> | <math>(A+B+C)+(C+D+E)+(F+G+H)=90</math> and <math>(B+C+D)+(E+F+G)=60</math> |
Revision as of 12:58, 4 June 2019
Problem 17
In the eight-term sequence , the value of is 5 and the sum of any three consecutive terms is 30. What is ?
Solution
We consider the sum and use the fact that , and hence .
Equating the two values we get for the sum, we get the answer .
Solution 2
Given that the sum of 3 consecutive terms is 30, we have and
It follows that because .
Subtracting, we have that .
Solution 2
We see that , and by substituting the given , we find that . Similarly, and .
Similarly, and , giving us . Since , .
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
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 |
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