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2010 AMC 12A Problem 10

Problem 10 of 25EasierAlgebra

The first four terms of an arithmetic sequence are p,p, 9,9, 3pq,3p-q, and 3p+q.3p+q. What is the 20102010th term of this sequence?

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Solution

Consecutive terms differ by a common difference d.d. From the last two terms, d=(3p+q)(3pq)=2q.d=(3p+q)-(3p-q)=2q. From the first two terms, 9p=d=2q,9-p=d=2q, and from the second and third, (3pq)9=d=2q.(3p-q)-9=d=2q. Solving this system gives p=5,p=5, q=2,q=2, and d=4.d=4. The 20102010th term is p+2009d=5+20094=8041. \begin{aligned} p+2009d &= 5+2009\cdot4 \\ &= 8041. \end{aligned} Thus, A is the correct answer.

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Concepts: arithmetic sequence · system of equations

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