2002 AMC 10P Problems/Problem 5

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, $a_{n+2}=\frac{1}{3}+a_{n+1}=a_{n+2}=\frac{1}{3}+\frac{1}{3}+a_n.$ 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.

\[\begin{tabular}{|c|c|c|c|c|c|}
 \hline 
 n & 1 & 2 & 3 & 4 & 5 \\ 
 \hline
 an & 1 & \frac{4}{3} & \frac{5/3} & 2 & \frac{7/3} \\
 \hline
\end{tabular}\] (Error compiling LaTeX. Unknown error_msg)

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 $\boxed{\textbf{(C) } 668}.$

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2002 AMC 10P Problems/Problem 5

Contents

Problem

Solution 1

Solution 2

See also