2008 AMC 12B Problems/Problem 12: Difference between revisions

Revision as of 10:37, 2 February 2015

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?

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

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.

$a_n = 2n - 1$

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

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

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

@@ Line 30: / Line 30: @@
 <math>a_1+a_2+\cdots+a_{2008}=2008^2</math> and <math>a_1+a_2+\cdots+a_{2007}=2007^2</math>
-Hence, by substitution, <math>a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{B}</math>
+Hence, by substitution, <math>a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{\textbf{B}}</math>
-(Solution by Bob_Smith)
 ==See Also==
 {{AMC12 box|year=2008|ab=B|num-b=11|num-a=13}}
 {{MAA Notice}}