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 1

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.

$a_n = 2n - 1$

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

Solution 2

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

Solution 3

Since the mean will be the sum of the first $n$ terms divided by $n$ , and said mean also equals $n$ , we know that the sum must be $n^2$ . This means the sequence must compute squares, and this is done by the odd integer sequence $1+3+5+7+\ldots$ . Therefore, we must find the 2008th odd number, which is found by $2n-1 \implies 2(2008)-1 = 4015$ , so the answer is $\boxed{\textbf{B}}$ ~stress-couture

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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2008 AMC 12B Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also