Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 10B Problems/Problem 15"

(→‎Solution 2)
m
(6 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
Let <math>S_n</math> be the sum of the first <math>n</math> term 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>?
+
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>
Line 18: Line 18:
 
Solving for <math>a</math>, we get that <math>a=1</math>.  
 
Solving for <math>a</math>, we get that <math>a=1</math>.  
  
−
Now, we proceed similar to Solution 1 and get that <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 3 (Quick Insight)==
 
==Solution 3 (Quick Insight)==
Line 35: Line 35:
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 
https://youtu.be/qkyRBpQHbOA
 
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}}

Revision as of 13:13, 21 June 2024

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

Suppose that the first number of the arithmetic sequence is $a$. We will try to compute the value of $S_{n}$. First, note that the sum of an arithmetic sequence is equal to the number of terms multiplied by the median of the sequence. The median of this sequence is equal to $a + n - 1$. Thus, the value of $S_{n}$ is $n(a + n - 1) = n^2 + n(a - 1)$. Then, \[\frac{S_{3n}}{S_{n}} = \frac{9n^2 + 3n(a - 1)}{n^2 + n(a - 1)} = 9 - \frac{6n(a-1)}{n^2 + n(a-1)}.\] Of course, for this value to be constant, $6n(a-1)$ must be $0$ for all values of $n$, and thus $a = 1$. Finally, we have $S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$.

~mathboy100

Solution 2

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 3 (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 $S_n = 20^2 = \boxed{\textbf{(D) } 400}$.

~numerophile

Video Solution (🚀 Solved in 4 min 🚀)

https://youtu.be/7ztNpblm2TY

~Education, the Study of Everything

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_15&oldid=220883"