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

Latest revision as of 01:02, 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

Remark 3.0.1

Hi! It me (Pinotation) again!

If you are confused about why $ a_1 \neq 3 $ or $ a_1 \neq 4 $ or something along those lines, we can prove by induction.

We have $S_n=\frac{n}{2}\big(2a_1+(n-1)\cdot 2\big)=n(a_1+n-1).$ Then $S_{3n}=3n(a_1+3n-1),$ so $\frac{S_{3n}}{S_n}=\frac{3(a_1+3n-1)}{a_1+n-1}.$

Base case $ n=1 $: $\frac{S_3}{S_1}=\frac{3(a_1+2)}{a_1}.$

Suppose for some $ n $ the ratio is independent of $ n $ and equal to a constant $ k $. Then $\frac{S_{3n}}{S_n}=\frac{3(a_1+3n-1)}{a_1+n-1}=k.$ For $ n+1 $ we must also have $\frac{S_{3(n+1)}}{S_{n+1}}=\frac{3(a_1+3n+2)}{a_1+n}=k.$ So $\frac{3(a_1+3n-1)}{a_1+n-1}=\frac{3(a_1+3n+2)}{a_1+n}.$ Cross multiplying gives $(a_1+3n-1)(a_1+n)=(a_1+3n+2)(a_1+n-1).$ Expanding, $a_1^2+4na_1+a_1- n -3 = a_1^2+4na_1+a_1+2n-2.$ Simplifying, $- n -3 = 2n -2,$ so $3n = -1.$ This is impossible for integer $ n $. The only way the equality can hold for all $ n $ is if the proportionality factor between numerator and denominator is fixed. Comparing coefficients in $3(a_1+3n-1)=C(a_1+n-1),$ we get $ C=9 $ and hence $3a_1-3=9a_1-9 \implies a_1=1.$

I hope this helped! If not, then you can check Solution 1, as it is practically the friendlier version LOL!

~Proof by 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 22: / Line 22: @@
 ~numerophile
-==Solution 3 (I didn't know Solution 1!)===
+==Solution 3 (I didn't know Solution 1!)==
 We have a slightly challenging problem :-(, but that's okay!
@@ Line 44: / Line 44: @@
 ===Remark 3.0.1===
-Hi! It me again!
+Hi! It me (Pinotation) again!
-If you are confused about why
+If you are confused about why \( a_1 \neq 3 \) or \( a_1 \neq 4 \) or something along those lines, we can prove by induction.
+We have
+<cmath>
+S_n=\frac{n}{2}\big(2a_1+(n-1)\cdot 2\big)=n(a_1+n-1).
+</cmath>
+Then
+<cmath>
+S_{3n}=3n(a_1+3n-1),
+</cmath>
+so
+<cmath>
+\frac{S_{3n}}{S_n}=\frac{3(a_1+3n-1)}{a_1+n-1}.
+</cmath>
+Base case \( n=1 \):
+<cmath>
+\frac{S_3}{S_1}=\frac{3(a_1+2)}{a_1}.
+</cmath>
+Suppose for some \( n \) the ratio is independent of \( n \) and equal to a constant \( k \). Then
+<cmath>
+\frac{S_{3n}}{S_n}=\frac{3(a_1+3n-1)}{a_1+n-1}=k.
+</cmath>
+For \( n+1 \) we must also have
+<cmath>
+\frac{S_{3(n+1)}}{S_{n+1}}=\frac{3(a_1+3n+2)}{a_1+n}=k.
+</cmath>
+So
+<cmath>
+\frac{3(a_1+3n-1)}{a_1+n-1}=\frac{3(a_1+3n+2)}{a_1+n}.
+</cmath>
+Cross multiplying gives
+<cmath>
+(a_1+3n-1)(a_1+n)=(a_1+3n+2)(a_1+n-1).
+</cmath>
+Expanding,
+<cmath>
+a_1^2+4na_1+a_1- n -3 = a_1^2+4na_1+a_1+2n-2.
+</cmath>
+Simplifying,
+<cmath>
+- n -3 = 2n -2,
+</cmath>
+so
+<cmath>
+n = -1.
+</cmath>
+This is impossible for integer \( n \). The only way the equality can hold for all \( n \) is if the proportionality factor between numerator and denominator is fixed. Comparing coefficients in
+<cmath>
+(a_1+3n-1)=C(a_1+n-1),
+</cmath>
+we get \( C=9 \) and hence
+<cmath>
+a_1-3=9a_1-9 \implies a_1=1.
+</cmath>
+I hope this helped! If not, then you can check Solution 1, as it is practically the friendlier version LOL!
+~Proof by Pinotation
 ==Video Solution (🚀 Solved in 5 min 🚀)==

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