Difference between revisions of "2022 AMC 10B Problems/Problem 15"
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==Problem== | ==Problem== | ||
− | Let <math>S_n</math> be the sum of the first <math>n</math> | + | Let <math>S_n</math> be the sum of the first <math>n</math> terms of an arithmetic sequence that has a common difference of <math>2</math>. The quotient <math>\frac{S_{3n}}{S_n}</math> does not depend on <math>n</math>. What is <math>S_{20}</math>? |
<math>\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420</math> | <math>\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420</math> | ||
==Solution 1== | ==Solution 1== | ||
+ | Let's say that our sequence is <cmath>a, a+2, a+4, a+6, a+8, a+10, \ldots.</cmath> | ||
+ | Then, since the value of n doesn't matter in the quotient <math>\frac{S_{3n}}{S_n}</math>, we can say that | ||
+ | <cmath>\frac{S_{3}}{S_1} = \frac{S_{6}}{S_2}.</cmath> | ||
+ | Simplifying, we get <math>\frac{3a+6}{a}=\frac{6a+30}{2a+2}</math>, from which <cmath>\frac{3a+6}{a}=\frac{3a+15}{a+1}.</cmath> <cmath>3a^2+9a+6=3a^2+15a</cmath> <cmath>6a=6</cmath> | ||
+ | Solving for <math>a</math>, we get that <math>a=1</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>. | |
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==Solution 2 (Quick Insight)== | ==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>. |
− | <math>\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9</math> | + | Since <math>\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9</math>, we have <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>. |
~numerophile | ~numerophile | ||
− | ==Solution | + | ==Video Solution (🚀 Solved in 5 min 🚀)== |
− | + | https://youtu.be/7ztNpblm2TY | |
− | |||
− | |||
− | + | ~Education, the Study of Everything | |
− | + | ==Video Solution By SpreadTheMathLove== | |
+ | https://www.youtube.com/watch?v=zHJJyMlH9DA | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/qkyRBpQHbOA | ||
+ | ==Video Solution by paixiao== | ||
+ | https://www.youtube.com/watch?v=4bzuoKi2Tes | ||
− | + | ==Video Solution by TheBeautyofMath== | |
− | + | https://youtu.be/Mi2AxPhnRno?t=1299 | |
== See Also == | == See Also == | ||
{{AMC10 box|year=2022|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2022|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:02, 10 November 2024
Contents
[hide]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 Then, since the value of n doesn't matter in the quotient , we can say that Simplifying, we get , from which 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 5 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 |
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