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

Revision as of 00:55, 10 September 2025

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

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 $\frac{3a+6}{a}=\frac{3a+15}{a+1}.$ $3a^2+9a+6=3a^2+15a$ $6a=6$ Solving for $a$ , we get that $a=1$ .

Since the sum of the first $n$ odd numbers is $n^2$ , $S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$ .

Solution 2 (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 $S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$ .

~numerophile

Solution 3 (I didn't know Solution 1!)=

We have a slightly challenging problem :-(, but that's okay!

$\textbf{If you didn't come up with the Solution 1 intuition}$ , then here is a more direct approach.

We want $\frac{S_{3n}}{S_n}$ to be a natural ($\frac{S_{3n}}{S_n} > 0$; basically a whole number $\neq 0$) number. Then only can the progression be incremental by 2.

Assume that $a_1 = 1$. Then, $a_2 = 3$, $a_3 = 5$, etc. We take the first term $n=1$, which is 1. Then $3n$; the next 3 terms will sum to 9. $\frac{9}{1} = 9$, and this is an

What about $a_1 = 0$. We immediately see that this would be undefined, so we cannot have $a_1 = 0$. So we say $a_1 = 2$. Then the sum for $n=1$ is 2, and for $3n$; it is simply 6. This is $6/2 = 3$, so it works!

Then, what about $a_2 = 3$? Then this means that the $n=2$ sum is 4, and the $3n$ sum is 36. This is $36/4 = 9$, and this is the same as the difference before, proving that $a_1 = 1$ is indeed correct.

And then? What about $a_2 = 4$? This means $n=2$ sum is 6, and the $3n$ sum is 42, but uh oh, the progression breaks! Therefore, the evens cannot work.

Therefore, $a_1 = 1$, $a_{20}$ is just the sum of the first 20 odd numbers, which is $\boxed{\textbf{(D) } 400}$

~Pinotation

Video Solution (🚀 Solved in 5 min 🚀)

https://youtu.be/7ztNpblm2TY

~Education, the Study of Everything

Video Solution By SpreadTheMathLove

https://www.youtube.com/watch?v=zHJJyMlH9DA

Video Solution by Interstigation

https://youtu.be/qkyRBpQHbOA

Video Solution by TheBeautyofMath

https://youtu.be/Mi2AxPhnRno?t=1299

@@ Line 34: / Line 34: @@
 What about \(a_1 = 0\). We immediately see that this would be undefined, so we cannot have \(a_1 = 0\). So we say \(a_1 = 2\). Then the sum for \(n=1\) is  2, and for \(3n\); it is simply 6. This is \(6/2 = 3\), so it works!
-Then, what about \(a_1 = 3\)? Then this means that the \(n=1\) sum is 3, and the \(3n\) sum is 15. This is \(15/3 = 5\), so this may also work.
+Then, what about \(a_2 = 3\)? Then this means that the \(n=2\) sum is 4, and the \(3n\) sum is 36. This is \(36/4 = 9\), and this is the same as the difference before, proving that \(a_1 = 1\) is indeed correct.
-And then? What about \(a_1 = 4\)? This means \(n=1\) sum is 4, and the \(3n\) sum is 18, but uh oh! \(18/4\) is not a natural number, so this doesn't work, and because the other odd case works, the even case cannot.
+And then? What about \(a_2 = 4\)? This means \(n=2\) sum is 6, and the \(3n\) sum is 42, but uh oh, the progression breaks! Therefore, the evens cannot work.
 Therefore, \(a_1 = 1\), \(a_{20}\) is just the sum of the first 20 odd numbers, which is <math>\boxed{\textbf{(D) } 400}</math>

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

Page

Toolbox

Search