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

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==Problem==
 
==Problem==
  
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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>?
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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>
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==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
  
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==Solution 2 (Quick Insight)==
+
==Solution 2==
 +
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>.
  
−
Recall that the sum of the first <math>n</math> odd numbers is <math>n^2</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>.
  
−
<math>\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9</math>. Thus <math>S_n = 20^2 = \fbox{D. 400}</math>
+
==Solution 3 (Quick Insight)==
 +
 
 +
Recall that the sum of the first <math>n</math> odd numbers is <math>n^2</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
 +
 +
==Video Solution (🚀 Solved in 4 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 14:42, 28 September 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 $S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$.

~numerophile

Video Solution (🚀 Solved in 4 min 🚀)

https://youtu.be/7ztNpblm2TY

~Education, the Study of Everything

Video Solution By 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

2022 AMC 10B (Problems • Answer Key • Resources)
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Problem 14 		Followed by
Problem 16
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All AMC 10 Problems and Solutions

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