Difference between revisions of "2008 AMC 10B Problems/Problem 11"

Revision as of 17:27, 25 December 2016

Problem

Suppose that $(u_n)$ is a sequence of real numbers satifying $u_{n+2}=2u_{n+1}+u_n$ ,

and that $u_3=9$ and $u_6=128$ . What is $u_5$ ?

$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 53\qquad\mathrm{(C)}\ 68\qquad\mathrm{(D)}\ 88\qquad\mathrm{(E)}\ 104$

Solution

Plugging in $n=3$ , we get

$128=2u_5+u_4.$

Plugging in $n=3$ , we get

$u_5=2u_4+9.$

This is simply a system of two equations with two unknowns. Substituting gives $128=5u_4+18 \Longrightarrow u_4=22$ , and $u_5=\frac{128-22}{2}=53 \Longrightarrow \textbf{(B)}$ .

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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.

@@ Line 7: / Line 7: @@
 ==Solution==
-Plugging in <math>n=4</math>, we get
+Plugging in <math>n=3</math>, we get
 <center><math>128=2u_5+u_4.</math></center>

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Difference between revisions of "2008 AMC 10B Problems/Problem 11"

Revision as of 17:27, 25 December 2016

Problem

Solution

See also