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
(→‎Solution 2)
 
(15 intermediate revisions by 8 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 11: Line 11:
 
~mathboy100
 
~mathboy100
  
−
==Solution 2 (Quick Insight)==
+
==Solution 2==
 +
We'll start with the ratio that term 3 ÷ term 1 = term 6 ÷ term 2
 +
 
 +
the sequence goes like: a, a+2, a+4, a+6, a+8, a+10...
 +
 
 +
term 3 ÷ term 1 = a+4 ÷ a
 +
 
 +
term 6 ÷ term 2 = a+10 ÷ a+2
 +
 
 +
a+4 ÷ a = a+10 ÷ a+2
 +
 
 +
(a+4)(a+2) = (a)(a+10)
 +
 
 +
a^2+6a+8 = a^2+10a
 +
 
 +
    6a+8=10a
 +
 
 +
        8=4a
 +
 
 +
        2=a
 +
 
 +
 
 +
the sequence is updated to 2,4,6,8,10,12...40
 +
 
 +
or 2(1+2+3+4+5+6...+20)
 +
 
 +
which is also 2(20 x 21)/2 or 20 x 21 and that is 420.
 +
 
 +
==Solution 3 (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>.
Line 19: Line 47:
 
~numerophile
 
~numerophile
  
−
==Solution 3==
+
==Video Solution (🚀 Solved in 4 min 🚀)==
−
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>.
 
−
 
 
−
We can simplify further to get <math>\frac{a+2}{a}=\frac{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>.
 
−
 
 
−
==Video Solution 1==
 
 
https://youtu.be/7ztNpblm2TY
 
https://youtu.be/7ztNpblm2TY
  
 
~Education, the Study of Everything
 
~Education, the Study of Everything
 +
 +
==Video Solution by Interstigation==
 +
https://youtu.be/qkyRBpQHbOA
 +
==Video Solution by paixiao==
 +
https://www.youtube.com/watch?v=4bzuoKi2Tes
  
 
== 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 13:34, 2 April 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

We'll start with the ratio that term 3 ÷ term 1 = term 6 ÷ term 2

the sequence goes like: a, a+2, a+4, a+6, a+8, a+10...

term 3 ÷ term 1 = a+4 ÷ a

term 6 ÷ term 2 = a+10 ÷ a+2

a+4 ÷ a = a+10 ÷ a+2

(a+4)(a+2) = (a)(a+10)

a^2+6a+8 = a^2+10a 

    6a+8=10a

       8=4a

       2=a


the sequence is updated to 2,4,6,8,10,12...40

or 2(1+2+3+4+5+6...+20)

which is also 2(20 x 21)/2 or 20 x 21 and that is 420.

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

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=217689"