Difference between revisions of "2008 AMC 10B Problems/Problem 11"
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==Solution== | ==Solution== | ||
− | + | If we plug in <math>n=4</math>, we get | |
<center><math>128=2u_5+u_4.</math></center> | <center><math>128=2u_5+u_4.</math></center> | ||
− | + | By plugging in <math>n=3</math>, we get | |
<center><math>u_5=2u_4+9.</math></center> | <center><math>u_5=2u_4+9.</math></center> | ||
− | This is | + | This is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives <math>128=5u_4+18 \Longrightarrow u_4=22</math>, therefore <math>u_5=\frac{128-22}{2}=53 \longrightarrow \textbf{\boxed{(B)}}</math>. |
==See also== | ==See also== |
Latest revision as of 16:15, 7 June 2024
Problem
Suppose that is a sequence of real numbers satifying
,
and that and
. What is
?
Solution
If we plug in , we get
![$128=2u_5+u_4.$](http://latex.artofproblemsolving.com/8/6/c/86c6e57b4980da77a3d343cef4e6c1825601f6df.png)
By plugging in , we get
![$u_5=2u_4+9.$](http://latex.artofproblemsolving.com/a/4/1/a413dedc5b1485079f3771cbdd231293ae79c5a0.png)
This is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives , therefore
.
See also
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.