Difference between revisions of "2008 AMC 10B Problems/Problem 13"
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==Solution 1== | ==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 <math>2008^{\text{th}}</math> term of the sequence is <math>2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{\ | + | 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{\textbf{(B) 4015}}</math> |
− | Note that <math>n^2</math> is the sum of the first n odd numbers. | + | Note that <math>n^2</math> is the sum of the first n odd integers. |
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+ | |||
+ | ==Solution 2== | ||
+ | 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{\textbf{(B) 4015}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let the mean be <math>\frac{(a)+(a+d)+(a+2d)+...+(a+(n-1)) \cdot d)}{n}</math> | ||
+ | <math>=\frac{n \cdot a}{n} + \frac{(1+2+3+...+(n-1)) \cdot d}{n}</math> | ||
+ | <math>=a + \frac{n \cdot (n-1) \cdot d}{2n}</math> | ||
+ | <math>=a+ \frac{(n-1) \cdot d}{2}</math> | ||
+ | |||
+ | Note that this is also equal to n | ||
+ | |||
+ | <math>a+ \frac{(n-1) \cdot d}{2}=n</math> | ||
+ | <math>\therefore 2a+ (n-1) \cdot d=2n</math> | ||
+ | |||
+ | 1st term + nth term <math>=2n=2 \cdot 2008=4016</math> | ||
+ | Now note that, from previous solutions, the first term is 1, hence the 2008th term is <math>4016-1=\boxed{\textbf{(B) 4015}}</math> | ||
+ | |||
+ | ~anshulb | ||
+ | |||
+ | ==Solution 4 (Using Answer Choices)== | ||
+ | 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{\textbf{(B) 4015}}</math> is the answer. | ||
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==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}} |
Latest revision as of 05:10, 1 October 2023
Contents
Problem
For each positive integer , the mean of the first
terms of a sequence is
. What is the
term of the sequence?
Solution 1
Since the mean of the first terms is
, the sum of the first
terms is
. Thus, the sum of the first
terms is
and the sum of the first
terms is
. Hence, the
term of the sequence is
Note that is the sum of the first n odd integers.
Solution 2
Let be the terms of the sequence. We know
, so we must have
. The sum of consecutive odd numbers down to
is a perfect square, if you don't believe me, try drawing squares with the sum, so
, so the answer is
.
Solution 3
Let the mean be
Note that this is also equal to n
1st term + nth term
Now note that, from previous solutions, the first term is 1, hence the 2008th term is
~anshulb
Solution 4 (Using Answer Choices)
From inspection, we see that the sum of the sequence is . We also notice that
is the sum of the first
odd integers. Because
is the only odd integer,
is the answer.
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
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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