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==Video Solution by Interstigation==
 
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==Video Solution by paixiao==
 
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==Video Solution by TheBeautyofMath==
 
==Video Solution by TheBeautyofMath==
 
https://youtu.be/Mi2AxPhnRno?t=1299
 
https://youtu.be/Mi2AxPhnRno?t=1299

Latest revision as of 09:39, 28 June 2025

Problem

Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence that has a common difference of $2$. The quotient $\frac{S_{3n}}{S_n}$ does not depend on $n$. What is $S_{20}$?

$\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420$

Solution 1

Let's say that our sequence is \[a, a+2, a+4, a+6, a+8, a+10, \ldots.\] Then, since the value of n doesn't matter in the quotient $\frac{S_{3n}}{S_n}$, we can say that \[\frac{S_{3}}{S_1} = \frac{S_{6}}{S_2}.\] Simplifying, we get $\frac{3a+6}{a}=\frac{6a+30}{2a+2}$, from which \[\frac{3a+6}{a}=\frac{3a+15}{a+1}.\] \[3a^2+9a+6=3a^2+15a\] \[6a=6\] Solving for $a$, we get that $a=1$.

Since the sum of the first $n$ odd numbers is $n^2$, $S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$.

Solution 2 (Quick Insight)

Recall that the sum of the first $n$ odd numbers is $n^2$.

Since $\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9$, we have $S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$.

~numerophile

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 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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