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2022 AMC 10B Problems/Problem 15

Revision as of 07:45, 19 November 2022 by Kavya.rajesh (talk | contribs)

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

Solution 3

Let's say that our sequence is $a, a+2, a+4, a+6, a+8, a+10...$

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 $(3a+6)/a$=$(6a+30)/(2a+2)$

We can simplify further to get $(a+2)/a$=$(a+5)/(a+1)$ Solving for $a$, we get that $a=1$. Now, we proceed similar to the previous solutions and get that $S_n = \fbox{D. 400}$

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

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