2022 AMC 10B Problems/Problem 15

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

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

See Also

2022 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 10 Problems and Solutions

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