2008 AMC 10B Problem 11 — Solution | LIVE by Po-Shen Loh

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2008 AMC 10B Problem 11

Attempt Problem 11 of the 2008 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

Suppose that $(u_n)$ is a sequence of real numbers satisfying $u_{n+2}=2u_{n+1}+u_n,$ and that $u_3=9$ and $u_6=128.$ What is $u_5?$

$40$

$53$

$68$

$88$

$104$

Answer: B

Concepts:recursion system of equations

Difficulty rating: 1140

Solution:

Using the recurrence, $u_5=2u_4+u_3=2u_4+9$ and $u_6=2u_5+u_4$ $=2(2u_4+9)+u_4$ $=5u_4+18.$

Setting $5u_4+18=128$ gives $u_4=22,$ so $u_5=2\cdot 22+9=53.$

Thus, the correct answer is B.

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