Difference between revisions of "2008 AMC 10B Problems/Problem 13"

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<math>\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}</math>
 
<math>\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}</math>
  
==Solution==
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==Solution 1==
Since the mean of the first <math>n</math> terms is <math>n</math>, the sum of the first <math>n</math> terms is <math>n^2</math>. Thus, the sum of the first <math>2007</math> terms is <math>2007^2</math> and the sum of the first <math>2008</math> terms is <math>2008^2</math>. Hence, the 2008th term is <math>2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{\text{(B)}}</math>
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Since the mean of the first <math>n</math> terms is <math>n</math>, the sum of the first <math>n</math> terms is <math>n^2</math>. Thus, the sum of the first <math>2007</math> terms is <math>2007^2</math> and the sum of the first <math>2008</math> terms is <math>2008^2</math>. Hence, the <math>2008^{\text{th}}</math> term of the sequence is <math>2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{\text{(B)}}</math>
  
Note that <math>n^2</math> is the sum of the first n odd numbers.
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Note that <math>n^2</math> is the sum of the first n odd integers.
  
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==Solution 2 (Using Answer Choices)==
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From inspection, we see that the sum of the sequence is <math>n^2</math>. We also notice that <math>n^2</math> is the sum of the first <math>n</math> odd integers. Because <math>4015</math> is the only odd integer, <math>\boxed{B}</math> is the answer.
  
==Solution 2(Using Answer Choices)==
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==Solution 3 (Basically Solution Two Just More Rigorous)==
From inspection, we see that the sum of the sequence is basically <math>n^2</math>. We also notice that <math>n^2</math> Is the sum of the first <math>n</math> ODD integers. Because <math>4015</math> is the only odd integer, <math>\boxed{B}</math> is the answer.
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Let <math>a_1, a_2, a_3, \cdots, a_n</math> be the terms of the sequence. We know <math>\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} = n</math>, so we must have <math>a_1 + a_2 + a_3 + \cdots + a_n = n^2</math>. The sum of consecutive odd numbers down to <math>1</math> is a perfect square, if you don't believe me, try drawing squares with the sum, so <math>a_1 = 1, a_2 = 3, a_3 = 5, \cdots , a_n = 2(n-1) + 1</math>, so the answer is <math>a_{2008} = 2(2007) + 1 = \boxed{\text{B}}</math>.
 
 
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2008|ab=B|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:20, 30 August 2022

Problem

For each positive integer $n$, the mean of the first $n$ terms of a sequence is $n$. What is the $2008^{\text{th}}$ term of the sequence?

$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$

Solution 1

Since the mean of the first $n$ terms is $n$, the sum of the first $n$ terms is $n^2$. Thus, the sum of the first $2007$ terms is $2007^2$ and the sum of the first $2008$ terms is $2008^2$. Hence, the $2008^{\text{th}}$ term of the sequence is $2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{\text{(B)}}$

Note that $n^2$ is the sum of the first n odd integers.

Solution 2 (Using Answer Choices)

From inspection, we see that the sum of the sequence is $n^2$. We also notice that $n^2$ is the sum of the first $n$ odd integers. Because $4015$ is the only odd integer, $\boxed{B}$ is the answer.

Solution 3 (Basically Solution Two Just More Rigorous)

Let $a_1, a_2, a_3, \cdots, a_n$ be the terms of the sequence. We know $\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} = n$, so we must have $a_1 + a_2 + a_3 + \cdots + a_n = n^2$. The sum of consecutive odd numbers down to $1$ is a perfect square, if you don't believe me, try drawing squares with the sum, so $a_1 = 1, a_2 = 3, a_3 = 5, \cdots , a_n = 2(n-1) + 1$, so the answer is $a_{2008} = 2(2007) + 1 = \boxed{\text{B}}$.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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 10 Problems and Solutions

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