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

Revision as of 19:49, 19 November 2022

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 $S_n = \fbox{D. 400}$

Video Solution 1

https://youtu.be/7ztNpblm2TY

~Education, the Study of Everything

@@ Line 4: / Line 4: @@
 <math>\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420</math>
-==Video Solution 1==
-https://youtu.be/7ztNpblm2TY
-~Education, the Study of Everything
 ==Solution 1==

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