For each positive integer $n$, the mean of the first $n$ terms of a sequence is $n$. What is the 2008th term of the sequence?
The mean of the first $n$ terms of a sequence is the sum of the first $n$ terms divided by $n$. So if the mean is $n$, then the sum of the first $n$ terms must be $n^2$. Let's call the sequence $a_1, a_2, a_3, \dots, a_n, \dots$. Then we have $a_1 + a_2 + a_3 + \cdots + a_n = n^2$ for each $n$. We want to find $a_{2008}$. To do that, we can subtract the sum of the first $2007$ terms from the sum of the first $2008$ terms, since that will give us the last term. So we have $a_{2008} = (a_1 + a_2 + a_3 + \cdots + a_{2008}) - (a_1 + a_2 + a_3 + \cdots + a_{2007}) = 2008^2 - 2007^2$. Now we just need to simplify this difference of squares. We can use the formula $(a^2 - b^2) = (a + b)(a - b)$ to factor it. So we have $a_{2008} = (2008 + 2007)(2008 - 2007) = 4015 \times 1 = 4015$. 