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

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2008 AMC 10B Problem 11

Below is the professionally curated solution for Problem 11 of the 2008 AMC 10B, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:recursion system of equations

Difficulty rating: 1140

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 $u_5?$

$40$

$53$

$68$

$88$

$104$

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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