2011 AMC 12A Problems/Problem 8: Difference between revisions
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== Solution == | == Solution == | ||
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> | 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> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=7|num-a=9|ab=A}} | {{AMC12 box|year=2011|num-b=7|num-a=9|ab=A}} | ||
Revision as of 19:10, 10 February 2011
Problem
In the eight term sequence 
, 
, 
, 
, 
, 
, 
, 
, the value of is 
and the sum of any three consecutive terms is 
. What is 
?
Solution
Let 
. Then from 
, we find that 
. From 
, we then get that 
. Continuing this pattern, we find 
, 
, 
, and finally 
. So 
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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 12 Problems and Solutions | |
