AMC12 2008 12B, 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?

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 $ $\fbox{4015}$.

Answer: 4015