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

Revision as of 19:55, 14 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, $a_{n+2}=\frac{1}{3}+a_{n+1}=a_{n+2}=\frac{1}{3}+\frac{1}{3}+a_n.$ 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. $\begin{tabular}{|c|c|c|c|c|c|} \hline n & 1 & 2 & 3 & 4 & 5 \\ \hline an & 1 & 4/3 & 5/3 & 2 & 7/3 \\ \hline \end{tabular}$

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

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

Revision as of 19:55, 14 July 2024

Contents

Problem

Solution 1

Solution 2

See also

@@ Line 26: / Line 26: @@
   n & 1 & 2 & 3 & 4 & 5 \\
   \hline
-  an & 1 & \frac{4}{3} & \frac{5/3} & 2 & \frac{7/3} \\
+  an & 1 & 4/3 & 5/3 & 2 & 7/3 \\
   \hline
 \end{tabular}