2022 AMC 10B Problems/Problem 15

Problem

Let $S_n$ be the sum of the first $n$ terms 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

We'll start with the ratio that term 3 ÷ term 1 = term 6 ÷ term 2

the sequence goes like: a, a+2, a+4, a+6, a+8, a+10...

term 3 ÷ term 1 = a+4 ÷ a

term 6 ÷ term 2 = a+10 ÷ a+2

a+4 ÷ a = a+10 ÷ a+2

(a+4)(a+2) = (a)(a+10)

a^2+6a+8 = a^2+10a

    6a+8=10a

       8=4a

2=a

the sequence is updated to 2,4,6,8,10,12...40

or 2(1+2+3+4+5+6...+20)

which is also 2(20 x 21)/2 or 20 x 21 and that is 420.

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

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
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