Difference between revisions of "2010 AMC 12B Problems/Problem 10"
(Made the solution much easier computationally.) |
|||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
− | We first sum the first <math>99</math> numbers: <math>\frac{99(100)}{2}=99\ | + | We first sum the first <math>99</math> numbers: <math>\frac{99(100)}{2}=99\cdot50</math>. Then, we know that the sum of the series is <math>99\cdot50+x</math>. There are <math>100</math> terms, so we can divide this sum by <math>100</math> and set it equal to <math>100x</math>: |
+ | <cmath>\frac{99\cdot50+x}{100}=100x \Rightarrow 99\cdot50=100^2x-x\Rightarrow \frac{99\cdot50}{100^2-1}=x</cmath> | ||
+ | Using difference of squares: | ||
+ | <cmath>x=\frac{99\cdot50}{101\cdot99}=\frac{50}{101}</cmath> | ||
+ | Thus, the answer is <math>\boxed{\text{B}}</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=9|num-a=11|ab=B}} | {{AMC12 box|year=2010|num-b=9|num-a=11|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:29, 17 January 2020
Problem 10
The average of the numbers and is . What is ?
Solution
We first sum the first numbers: . Then, we know that the sum of the series is . There are terms, so we can divide this sum by and set it equal to : Using difference of squares: Thus, the answer is .
See also
2010 AMC 12B (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.