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"

m (→‎Video Solution 1)
m
Line 7: Line 7:
 
==Solution 1==
 
==Solution 1==
  
−
Suppose that the first number of the arithmetic sequence is <math>a</math>. We will try to compute the value of <math>S_{n}</math>. 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 <math>a + n - 1</math>. Thus, the value of <math>S_{n}</math> is <math>n(a + n - 1) = n^2 + n(a - 1)</math>. Then, <cmath>\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)}.</cmath> Of course, for this value to be constant, <math>6n(a-1)</math> must be <math>0</math> for all values of <math>n</math>, and thus <math>a = 1</math>. Finally, the value of <math>S_{20}</math> is <math>20^2 = \fbox{D. 400}</math>
+
Suppose that the first number of the arithmetic sequence is <math>a</math>. We will try to compute the value of <math>S_{n}</math>. 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 <math>a + n - 1</math>. Thus, the value of <math>S_{n}</math> is <math>n(a + n - 1) = n^2 + n(a - 1)</math>. Then, <cmath>\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)}.</cmath> Of course, for this value to be constant, <math>6n(a-1)</math> must be <math>0</math> for all values of <math>n</math>, and thus <math>a = 1</math>. Finally, we have <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>.
  
 
~mathboy100
 
~mathboy100
Line 13: Line 13:
 
==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>. Thus <math>S_n = 20^2 = \fbox{D. 400}</math>
+
Since <math>\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9</math>, we have <math>S_n = 20^2 = \boxed{\textbf{(D) } 400}</math>.
  
 
~numerophile
 
~numerophile
  
 
==Solution 3==
 
==Solution 3==
−
Let's say that our sequence is <math>a, a+2, a+4, a+6, a+8, a+10...</math>
+
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
 
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>.
  
−
<math>\frac{S_{3}}{S_1}</math> = <math>\frac{S_{6}}{S_2}</math>
+
We can simplify further to get <math>\frac{a+2}{a}=\frac{a+5}{a+1}</math>.
−
 
 
−
Simplifying, we get <math>(3a+6)/a</math>=<math>(6a+30)/(2a+2)</math>
 
  
−
We can simplify further to get <math>(a+2)/a</math>=<math>(a+5)/(a+1)</math>
+
Solving for <math>a</math>, we get that <math>a=1</math>. Now, we proceed similar to the previous solutions and get that <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>.
−
Solving for <math>a</math>, we get that <math>a=1</math>. Now, we proceed similar to the previous solutions and get that <math>S_n = \fbox{D. 400}</math>
 
  
 
==Video Solution 1==
 
==Video Solution 1==

Revision as of 06:53, 28 November 2022

Problem

Let $S_n$ be the sum of the first $n$ term 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 (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

Solution 3

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 $\frac{3a+6}{a}=\frac{6a+30}{2a+2}$.

We can simplify further to get $\frac{a+2}{a}=\frac{a+5}{a+1}$.

Solving for $a$, we get that $a=1$. Now, we proceed similar to the previous solutions and get that $S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$.

Video Solution 1

https://youtu.be/7ztNpblm2TY

~Education, the Study of Everything

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=183228"