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

Revision as of 06:54, 28 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, 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 $\frac{a+2}{a}=\frac{a+5}{a+1}.$ Solving for $a$ , we get that $a=1$ . Now, we proceed similar to Solution 1 and get that $S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$ .

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 1

https://youtu.be/7ztNpblm2TY

~Education, the Study of Everything

@@ Line 11: / Line 11: @@
 ~mathboy100
-==Solution 2 (Quick Insight)==
+==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
+<cmath>\frac{S_{3}}{S_1} = \frac{S_{6}}{S_2}.</cmath>
+Simplifying, we get <math>\frac{3a+6}{a}=\frac{6a+30}{2a+2}</math>, from which <cmath>\frac{a+2}{a}=\frac{a+5}{a+1}.</cmath>
+Solving for <math>a</math>, we get that <math>a=1</math>. Now, we proceed similar to Solution 1 and get that <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>.
+==Solution 3 (Quick Insight)==
 Recall that the sum of the first <math>n</math> odd numbers is <math>n^2</math>.
@@ Line 18: / Line 25: @@
 ~numerophile
-==Solution 3==
-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
-<cmath>\frac{S_{3}}{S_1} = \frac{S_{6}}{S_2}.</cmath>
-Simplifying, we get <math>\frac{3a+6}{a}=\frac{6a+30}{2a+2}</math>.
-We can simplify further to get <math>\frac{a+2}{a}=\frac{a+5}{a+1}</math>.
-Solving for <math>a</math>, we get that <math>a=1</math>. Now, we proceed similar to the previous solutions and get that <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>.
 ==Video Solution 1==

2022 AMC 10B (Problems • Answer Key • Resources)
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