Difference between revisions of "2008 AMC 12B Problems/Problem 12"
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| − | ==Problem | + | ==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? | ||
<math>\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064</math> | <math>\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Letting <math>S_n</math> be the nth partial sum of the sequence: | Letting <math>S_n</math> be the nth partial sum of the sequence: | ||
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| − | == | + | ==Solution 2== |
Letting the sum of the sequence equal <math>a_1+a_2+\cdots+a_n</math> yields the following two equations: | 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_{2007}}{2007}=2007</math>. | <math>\frac{a_1+a_2+\cdots+a_{2007}}{2007}=2007</math>. | ||
| − | + | Therefore: | |
<math>a_1+a_2+\cdots+a_{2008}=2008^2</math> and <math>a_1+a_2+\cdots+a_{2007}=2007^2</math> | <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{\textbf{B}}</math> | |
| + | |||
| + | ==Solution 3== | ||
| + | Since the mean will be the sum of the first <math>n</math> terms divided by <math>n</math>, and said mean also equals <math>n</math>, we know that the sum must be <math>n^2</math>. This means the sequence must compute squares, and this is done by the odd integer sequence <math>1+3+5+7+\ldots</math>. Therefore, we must find the 2008th odd number, which is found by <math>2n-1 \implies 2(2008)-1 = 4015</math>, so the answer is <math>\boxed{\textbf{B}}</math> | ||
| + | ~stress-couture | ||
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 12:05, 21 May 2025
Problem
For each positive integer 
, the mean of the first terms of a sequence is . What is the 
th term of the sequence?
Solution 1
Letting 
be the nth partial sum of the sequence:
The only possible sequence with this result is the sequence of odd integers.
Solution 2
Letting the sum of the sequence equal 
yields the following two equations:
and
.
Therefore:
and 
Hence, by substitution, 
Solution 3
Since the mean will be the sum of the first terms divided by , and said mean also equals , we know that the sum must be 
. This means the sequence must compute squares, and this is done by the odd integer sequence 
. Therefore, we must find the 2008th odd number, which is found by 
, so the answer is 
~stress-couture
See Also
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
