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, the value of $S_{20}$ is $20^2 = \fbox{D. 400}$

~mathboy100

Solution 2 (Quick Insight)

Recall that the sum of the first $n$ odd numbers is $n^2$ .

$\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9$ . Thus $S_n = 20^2 = \fbox{D. 400}$

~numerophile

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2022 AMC 10B Problems/Problem 15

Contents

Problem

Solution 1

Solution 2 (Quick Insight)

See Also