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2008 AMC 12B Problems/Problem 12

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Problem 12

For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008$ th term of the sequence?

$\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032...$

Solution

Letting $S_n$ be the nth partial sum of the sequence:

$\frac{S_n}{n} = n$

The only possible sequence with this result is the sequence of odd integers.

$a_{2008} = 2(2008) - 1 = 4015 \Rightarrow \textbf{(B)}$

See Also

2008 AMC 12B (Problems • 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

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Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

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