Difference between revisions of "2010 AMC 12B Problems/Problem 10"
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Revision as of 00:15, 22 December 2020
- The following problem is from both the 2010 AMC 12B #10 and 2010 AMC 10B #14, so both problems redirect to this page.
Contents
[hide]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 .
Video Solution
https://youtu.be/vYXz4wStBUU?t=413
~IceMatrix
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 |
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.