2004 AMC 10B Problem 19

Below is the professionally curated solution for Problem 19 of the 2004 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10B solutions, or check the answer key.

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Concepts:recursion arithmetic sequence pattern recognition

Difficulty rating: 1460

19.

In the sequence $2001, 2002, 2003, \ldots,$ each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000.$ What is the $2004$ th term in this sequence?

$-2004$

$-2$

$0$

$4003$

$6007$

Solution:

The recurrence $a_{k+1} = a_{k-2} + a_{k-1} - a_k$ gives $a_{k+1} - a_{k-1} = -(a_k - a_{k-2}).$ The sequence begins $2001, 2002, 2003, 2000, 2005, 1998, \ldots$

So the even-position terms form the arithmetic sequence $2002, 2000, 1998, \ldots$ with common difference $-2.$ The $2004$ th term is its $1002$ nd term, $2002 + 1001(-2) = 0.$

Thus, the correct answer is C.

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