Difference between revisions of "2008 AMC 12B Problems/Problem 12"

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==Problem 12==
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==Problem==
 
For each positive integer <math>n</math>, the mean of the first <math>n</math> terms of a sequence is <math>n</math>. What is the <math>2008</math>th term of the sequence?
 
For each positive integer <math>n</math>, the mean of the first <math>n</math> terms of a sequence is <math>n</math>. What is the <math>2008</math>th term of the sequence?
  
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<math>a_{2008} = 2(2008) - 1 = 4015 \Rightarrow \textbf{(B)}</math>
 
<math>a_{2008} = 2(2008) - 1 = 4015 \Rightarrow \textbf{(B)}</math>
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==Alternate Solution==
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Letting the sum of the sequence equal <math>a_1+a_2+\cdots+a_n</math> yields the following two equations:
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<math>\frac{a_1+a_2+\cdots+a_{2008}}{2008}=2008</math> and
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<math>\frac{a_1+a_2+\cdots+a_{2007}}{2007}=2007</math>.
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Therefore:
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<math>a_1+a_2+\cdots+a_{2008}=2008^2</math> and <math>a_1+a_2+\cdots+a_{2007}=2007^2</math>
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Hence, by substitution, <math>a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{\textbf{B}}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=B|num-b=11|num-a=13}}
 
{{AMC12 box|year=2008|ab=B|num-b=11|num-a=13}}
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{{MAA Notice}}

Latest revision as of 12:49, 15 February 2021

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

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
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

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