Difference between revisions of "2013 AMC 12A Problems/Problem 7"
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<math>S_4 = S_6 - S_5 = 26 - 16 = 10</math> | <math>S_4 = S_6 - S_5 = 26 - 16 = 10</math> | ||
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+ | Therefore, the answer is <math>\boxed{\textbf{(C) }{10}}</math> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=A|num-b=6|num-a=8}} | {{AMC12 box|year=2013|ab=A|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:06, 25 January 2014
Problem
The sequence has the property that every term beginning with the third is the sum of the previous two. That is,
Suppose that
and
. What is
?
Solution
,
Therefore, the answer is
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.