2008 AMC 12B Problems/Problem 12

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?

$\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064$

Solution

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

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

$S_n = n^2$

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

$a_n = 2n - 1$

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

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 12 Problems and Solutions

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