# 2008 AMC 12B Problems/Problem 12

## Problem

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

## Alternate Solution

Letting the sum of the sequence equal $a_1+a_2+\cdots+a_n$ yields the following two equations: $\frac{a_1+a_2+\cdots+a_{2008}}{2008}=2008$ and $\frac{a_1+a_2+\cdots+a_{2007}}{2007}=2007$.

Therefore: $a_1+a_2+\cdots+a_{2008}=2008^2$ and $a_1+a_2+\cdots+a_{2007}=2007^2$

Hence, by substitution, $a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{\textbf{B}}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 