2022 AMC 10B Problems/Problem 15: Difference between revisions

Latest revision as of 00:59, 3 November 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 $n^2$ , $S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$ .

Note: you could also plug in the formulas for $\frac{S_{3n}}{S_{n}}$ and simplify, getting $3+ \frac{6n}{a+n-1}$ You would then find a= $1$ ~megacleverstarfish15

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 arithmetic sequence, and therefore works!

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

Solution 4 (Answer Choices)

Let $a$ be the first element. Then, $S_{20}=20(a+19)$ . Let it equal to all the answer choices respectively, we get $a=-2, -1, 0, 1, 2$ . We test $\frac{S_{3n}}{S_n}$ for each $a$ , using $n=1, 2$ . We can see that for $a=-2, -1, 0$ there exists $n$ to let $S_n=0$ , which implies they are invalid. For $a=2$ we can also see it's invalid, since $\frac{S_3}{S_1}=6\neq7=\frac{S_6}{S_2}$ . Therefore, $a=1$ , and $\boxed{\textbf{(D) } 400}$ is the correct choice.

~metrixgo

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 6: / Line 6: @@
 ==Solution 1==
-Suppose that the first number of the arithmetic sequence is <math>a</math>. We will try to compute the value of <math>S_{n}</math>. 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 <math>a + n - 1</math>. Thus, the value of <math>S_{n}</math> is <math>n(a + n - 1) = n^2 + n(a - 1)</math>. Then, <cmath>\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)}.</cmath> Of course, for this value to be constant, <math>6n(a-1)</math> must be <math>0</math> for all values of <math>n</math>, and thus <math>a = 1</math>. Finally, we have <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>.
-~mathboy100
-==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
@@ Line 20: / Line 14: @@
 Since the sum of the first <math>n</math> odd numbers is <math>n^2</math>, <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>.
-==Solution 3 (Quick Insight)==
+Note: you could also plug in the formulas for <math>\frac{S_{3n}}{S_{n}}</math> and simplify, getting <math>3+ \frac{6n}{a+n-1}</math> You would then find a=<math>1</math>
+~megacleverstarfish15
+==Solution 2 (Quick Insight)==
 Recall that the sum of the first <math>n</math> odd numbers is <math>n^2</math>.
-Since <math>\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9</math>, we have <math>S_n = 20^2 = \boxed{\textbf{(D) } 400}</math>.
+Since <math>\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9</math>, we have <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>.
 ~numerophile
-==Video Solution (🚀 Solved in 4 min 🚀)==
+==Solution 3 (I didn't know Solution 1!)==
+We have a slightly challenging problem :-(, but that's okay!
+<math>\textbf{If you didn't come up with the Solution 1 intuition}</math>, 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 arithmetic sequence, and therefore works!
+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 <math>\boxed{\textbf{(D) } 400}</math>
+~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
+<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
+==Solution 4 (Answer Choices)==
+Let <math>a</math> be the first element. Then, <math>S_{20}=20(a+19)</math>. Let it equal to all the answer choices respectively, we get <math>a=-2, -1, 0, 1, 2</math>. We test <math>\frac{S_{3n}}{S_n}</math> for each <math>a</math>, using <math>n=1, 2</math>. We can see that for <math>a=-2, -1, 0</math> there exists <math>n</math> to let <math>S_n=0</math>, which implies they are invalid. For <math>a=2</math> we can also see it's invalid, since <math>\frac{S_3}{S_1}=6\neq7=\frac{S_6}{S_2}</math>. Therefore, <math>a=1</math>, and <math>\boxed{\textbf{(D) } 400}</math> is the correct choice.
+~metrixgo
+==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 paixiao==
-https://www.youtube.com/watch?v=4bzuoKi2Tes
 ==Video Solution by TheBeautyofMath==
 https://youtu.be/Mi2AxPhnRno?t=1299

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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