Difference between revisions of "2022 AMC 10B Problems/Problem 15"
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Since the sum of the first <math>n</math> odd numbers is <math>n^2</math>, <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>. | Since the sum of the first <math>n</math> odd numbers is <math>n^2</math>, <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>. | ||
| − | ==Solution | + | ==Solution 2 (Quick Insight)== |
Recall that the sum of the first <math>n</math> odd numbers is <math>n^2</math>. | Recall that the sum of the first <math>n</math> odd numbers is <math>n^2</math>. | ||
Revision as of 11:26, 3 November 2024
Contents
Problem
Let 
be the sum of the first 
terms of an arithmetic sequence that has a common difference of 
. The quotient 
does not depend on . What is 
?
Solution 1
Let's say that our sequence is ![\[a, a+2, a+4, a+6, a+8, a+10, \ldots.\]](http://latex.artofproblemsolving.com/6/9/4/6949385de927daa2c0a1ca2865fa033d023d5684.png)
Then, since the value of n doesn't matter in the quotient , we can say that
![\[\frac{S_{3}}{S_1} = \frac{S_{6}}{S_2}.\]](http://latex.artofproblemsolving.com/2/1/1/211d4a410ea2f4e0fbb5ba5b4cfd339a58f3b8f1.png)
Simplifying, we get 
, from which ![\[\frac{3a+6}{a}=\frac{3a+15}{a+1}.\]](http://latex.artofproblemsolving.com/7/b/5/7b50f89c009267fd328dfb139693cfd6a8a6d21f.png)
![\[3a^2+9a+6=3a^2+15a\]](http://latex.artofproblemsolving.com/8/6/7/867df6bd88334ac5b0f1459607cc2e620226e9d2.png)
![\[6a=6\]](http://latex.artofproblemsolving.com/c/8/8/c88b943f424efaec2dad7f36bbe4f1effae81a16.png)
Solving for 
, we get that 
.
Since the sum of the first odd numbers is 
, 
.
Solution 2 (Quick Insight)
Recall that the sum of the first odd numbers is .
Since 
, we have .
~numerophile
Video Solution (🚀 Solved in 4 min 🚀)
~Education, the Study of Everything
Video Solution By SpreadTheMathLove
https://www.youtube.com/watch?v=zHJJyMlH9DA
Video Solution by Interstigation
Video Solution by paixiao
https://www.youtube.com/watch?v=4bzuoKi2Tes
Video Solution by TheBeautyofMath
https://youtu.be/Mi2AxPhnRno?t=1299
See Also
| 2022 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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. 
