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

Revision as of 16:35, 10 August 2008

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

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

(A) 40 (B) 53 (C) 68 (D) 88 (E) 104}}

{{We konw that $u_6=128$ , so we plug in $n=4$ to get $128=2u_5+u_4$ . We plug in $n=3$ to get

$u_5=2u_4+9$ . Substituting gives

$128=5u_4+18 \rightarrow u_4=22$

This gives $u_5=\frac{128-22}{2}=53$ .

Answer B is the correct answer

NOTE: This is my (BOGTRO) solution, not the official one,

and should be ignored in view of a better solution.}}

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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

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

@@ Line 1: / Line 1: @@
-==Problem==
-Postman Pete has a pedometer to count his steps. The pedometer records up to 99999 steps, then flips over to 00000 on the next step. Pete plans to determine his mileae for a year. On January 1 Pete sets the pedometer to 00000. During the year, the pedometer flips from 99999 to 00000 forty-four
-times. On December 31 the pedometer reads 50000. Pete takes 1800 steps per mile. Which of the following is closest to the number of miles Pete walked during the year?
-(A) 2500 (B) 3000 (C) 3500 (D) 4000 (E) 4500
 ==Problem==
 {{Suppose that <math>(u_n)</math> is a serquence of real numbers satifying <math>u_{n+2}=2u_{n+1}+u_n</math>,