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Difference between revisions of "2002 AMC 10P Problems/Problem 5"

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</math>
 
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== Solution 1==
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== Solution 1 ==
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The recursive rule is equal to <math>a_{n+1}=\frac{1}{3}+a_{n}</math> for all <math>n \geq 1.</math> By recursion, <math>a_{n+2}=\frac{1}{3}+a_{n+1}=\frac{1}{3}+\frac{1}{3}+a_n=\frac{1}{3}(2)+a_n.</math> If we set <math>n=1</math> and repeat this process <math>2001</math> times, we will get <math>a_{2001+1}=a_{2002}=\frac{1}{3}(2001) + a_1=667+1=668.</math>
  
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Thus, our answer is <math>\boxed{\textbf{(C) } 668}.</math>
  
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== Solution 2 ==
 +
Find the first few terms of the sequence and find a pattern.
 +
<cmath>
 +
\begin{tabular}{|c|c|c|c|c|c|}
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\hline
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n & 1 & 2 & 3 & 4 & 5 \\
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\hline
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an & 1 & 4/3 & 5/3 & 2 & 7/3 \\
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\hline
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\end{tabular}
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</cmath>
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 +
From here, we can deduce that <math>a_n=\frac{2}{3}+\frac{1}{3}n.</math> Plugging in <math>n=2002,</math> <math>a_{2002}=\frac{2}{3}+\frac{2002}{3}=668.</math>
 +
 +
Thus, our answer is <math>\boxed{\textbf{(C) } 668}.</math>
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2002|ab=P|num-b=4|num-a=6}}
 
{{AMC10 box|year=2002|ab=P|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:12, 15 July 2024

Problem

Let $(a_n)_{n \geq 1}$ be a sequence such that $a_1 = 1$ and $3a_{n+1} - 3a_n = 1$ for all $n \geq 1.$ Find $a_{2002}.$

$\text{(A) }666 \qquad \text{(B) }667 \qquad \text{(C) }668 \qquad \text{(D) }669 \qquad \text{(E) }670$

Solution 1

The recursive rule is equal to $a_{n+1}=\frac{1}{3}+a_{n}$ for all $n \geq 1.$ By recursion, $a_{n+2}=\frac{1}{3}+a_{n+1}=\frac{1}{3}+\frac{1}{3}+a_n=\frac{1}{3}(2)+a_n.$ If we set $n=1$ and repeat this process $2001$ times, we will get $a_{2001+1}=a_{2002}=\frac{1}{3}(2001) + a_1=667+1=668.$

Thus, our answer is $\boxed{\textbf{(C) } 668}.$

Solution 2

Find the first few terms of the sequence and find a pattern. \[\begin{tabular}{|c|c|c|c|c|c|} \hline n & 1 & 2 & 3 & 4 & 5 \\ \hline an & 1 & 4/3 & 5/3 & 2 & 7/3 \\ \hline \end{tabular}\]

From here, we can deduce that $a_n=\frac{2}{3}+\frac{1}{3}n.$ Plugging in $n=2002,$ $a_{2002}=\frac{2}{3}+\frac{2002}{3}=668.$

Thus, our answer is $\boxed{\textbf{(C) } 668}.$

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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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