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

Latest revision as of 14:02, 10 November 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

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 paixiao

https://www.youtube.com/watch?v=4bzuoKi2Tes

Video Solution by TheBeautyofMath

https://youtu.be/Mi2AxPhnRno?t=1299

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.

@@ Line 6: / Line 6: @@
 ==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, we have <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>.
-~mathboy100
-==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
@@ Line 20: / Line 14: @@
 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 2 (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_n = 20^2 = \boxed{\textbf{(D) } 400}</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
-==Video Solution (🚀 Solved in 4 min 🚀)==
+==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

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