Difference between revisions of "2008 AMC 12B Problems/Problem 12"
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<math>a_1+a_2+\cdots+a_{2008}=2008^2</math> and <math>a_1+a_2+\cdots+a_{2007}=2007^2</math> | <math>a_1+a_2+\cdots+a_{2008}=2008^2</math> and <math>a_1+a_2+\cdots+a_{2007}=2007^2</math> | ||
− | Hence, by substitution, <math>a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{B}</math> | + | Hence, by substitution, <math>a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{\textbf{B}}</math> |
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=11|num-a=13}} | {{AMC12 box|year=2008|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:37, 2 February 2015
Problem 12
For each positive integer , the mean of the first
terms of a sequence is
. What is the
th term of the sequence?
Solution
Letting be the nth partial sum of the sequence:
The only possible sequence with this result is the sequence of odd integers.
Alternate Solution
Letting the sum of the sequence equal yields the following two equations:
and
.
Therefore:
and
Hence, by substitution,
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.