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Difference between revisions of "2022 AMC 10A Problems/Problem 20"

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<math>\textbf{(A) } 190 \qquad \textbf{(B) } 194 \qquad \textbf{(C) } 198 \qquad \textbf{(D) } 202 \qquad \textbf{(E) } 206</math>
 
<math>\textbf{(A) } 190 \qquad \textbf{(B) } 194 \qquad \textbf{(C) } 198 \qquad \textbf{(D) } 202 \qquad \textbf{(E) } 206</math>
  
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==Solution 1==
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==Solution 1 (Algebra)==
 
Let the arithmetic sequence be <math>a,a+d,a+2d,a+3d</math> and the geometric sequence be <math>b,br,br^2,br^3.</math>
 
Let the arithmetic sequence be <math>a,a+d,a+2d,a+3d</math> and the geometric sequence be <math>b,br,br^2,br^3.</math>
  
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\end{align*}</cmath>
 
\end{align*}</cmath>
 
Subtract these results, we get <cmath>b(r-1)^2=28.</cmath>
 
Subtract these results, we get <cmath>b(r-1)^2=28.</cmath>
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Note that either <math>(r-1)^2=1</math> or <math>(r-1)^2=4.</math> We proceed with casework:
  
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Note that either <math>b=28</math> or <math>b=7.</math> We proceed with casework:
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* If <math>(r-1)^2=1,</math> then <math>r=2,b=28,a=29,</math> and <math>d=-25.</math> The arithmetic sequence is <math>29,4,-21,-46,</math> arriving at a contradiction.
  
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* If <math>b=28,</math> then <math>r=2,a=29,</math> and <math>d=25.</math> The arithmetic sequence is <math>29,4,-21,-46,</math> arriving at a contradiction.
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* If <math>(r-1)^2=4,</math> then <math>r=3,b=7,a=50,</math> and <math>d=-11.</math> The arithmetic sequence is <math>50,39,28,17,</math> and the geometric sequence is <math>7,21,63,189.</math> This case is valid.
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* If <math>b=7,</math> then <math>r=3,a=50,</math> and <math>d=-11.</math> The arithmetic sequence is <math>50,39,28,17,</math> and the geometric sequence is <math>7,21,63,189.</math> This case is valid.
 
  
 
Therefore, The answer is <math>a+3d+br^3=17+189=\boxed{\textbf{(E) } 206}.</math>
 
Therefore, The answer is <math>a+3d+br^3=17+189=\boxed{\textbf{(E) } 206}.</math>
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~mathboy282 ~MRENTHUSIASM
 
~mathboy282 ~MRENTHUSIASM
  
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==Solution 2==
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==Solution 2 (Answer Choices)==
  
 
Start similarly to Solution 1 and deduce the three equations
 
Start similarly to Solution 1 and deduce the three equations
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We are looking for <math>a+3d+27b</math>. If <math>\textbf{(A)}</math> were the answer, then we know that <math>a</math> would have to be divisible by <math>3</math> and <math>b</math> would equal <math>6</math>. Looking at our second equation, if this were the case, then <math>d</math> would also have to be divisible by <math>3</math>. However, this contradicts the third equation, as all variables are divisible by <math>3</math>, but their sum isn't. So, <math>\boxed{\textbf{(E) } 206}</math> is our answer.
 
We are looking for <math>a+3d+27b</math>. If <math>\textbf{(A)}</math> were the answer, then we know that <math>a</math> would have to be divisible by <math>3</math> and <math>b</math> would equal <math>6</math>. Looking at our second equation, if this were the case, then <math>d</math> would also have to be divisible by <math>3</math>. However, this contradicts the third equation, as all variables are divisible by <math>3</math>, but their sum isn't. So, <math>\boxed{\textbf{(E) } 206}</math> is our answer.
 +
 +
==Video Solution (🚀Just 4 min!🚀)==
 +
https://youtu.be/P13VHc26Sgo
 +
~<i>Education, the Study of Everything </i>
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Latest revision as of 19:58, 31 October 2024

Problem

A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are $57$, $60$, and $91$. What is the fourth term of this sequence?

$\textbf{(A) } 190 \qquad \textbf{(B) } 194 \qquad \textbf{(C) } 198 \qquad \textbf{(D) } 202 \qquad \textbf{(E) } 206$

Solution 1 (Algebra)

Let the arithmetic sequence be $a,a+d,a+2d,a+3d$ and the geometric sequence be $b,br,br^2,br^3.$

We are given that \begin{align*} a+b&=57, \\ a+d+br&=60, \\ a+2d+br^2&=91, \end{align*} and we wish to find $a+3d+br^3.$

Subtracting the first equation from the second and the second equation from the third, we get \begin{align*} d+b(r-1)&=3, \\ d+br(r-1)&=31. \end{align*} Subtract these results, we get \[b(r-1)^2=28.\] Note that either $(r-1)^2=1$ or $(r-1)^2=4.$ We proceed with casework:

  • If $(r-1)^2=1,$ then $r=2,b=28,a=29,$ and $d=-25.$ The arithmetic sequence is $29,4,-21,-46,$ arriving at a contradiction.
  • If $(r-1)^2=4,$ then $r=3,b=7,a=50,$ and $d=-11.$ The arithmetic sequence is $50,39,28,17,$ and the geometric sequence is $7,21,63,189.$ This case is valid.

Therefore, The answer is $a+3d+br^3=17+189=\boxed{\textbf{(E) } 206}.$

~mathboy282 ~MRENTHUSIASM

Solution 2 (Answer Choices)

Start similarly to Solution 1 and deduce the three equations \begin{align*} a+b&=57, \\ a+d+br&=60, \\ a+2d+br^2&=91. \end{align*} Then, add the last two equations and take away the first equation to get $a+3d+br^2+br-b=94$ We can solve for this in terms of what we want: $a+3d=-br^2-br+b+94$.

We're looking for $a+3d+br^3$. We can substitute our value of $a+3d$ in here to get \[br^3-br^2-br+b+94=b(r+1)(r-1)(r-1)+94.\] Since our sequence only has positive integers we can now check by the answer choices. For each answer choice, we can subtract $94$ and factor it to see if it has a perfect square factor and at least one other factor and those should differ by $2$. \begin{alignat*}{8} \textbf{(A)} \ 190-94&=96&&=2^5\cdot3, \\ \textbf{(B)} \ 194-94&=100&&=2^2\cdot5^2, \\ \textbf{(C)} \ 198-94&=104&&=2^3\cdot13, \\ \textbf{(D)} \ 202-94&=108&&=2^2\cdot3^3, \\ \textbf{(E)} \ 206-94&=112&&=2^4\cdot7. \end{alignat*} From this, the only possible answer choices are $\textbf{(A)}$ and $\textbf{(E)}$, where $r=3$. To solve for $b$, we look back to the given equations above.

We are looking for $a+3d+27b$. If $\textbf{(A)}$ were the answer, then we know that $a$ would have to be divisible by $3$ and $b$ would equal $6$. Looking at our second equation, if this were the case, then $d$ would also have to be divisible by $3$. However, this contradicts the third equation, as all variables are divisible by $3$, but their sum isn't. So, $\boxed{\textbf{(E) } 206}$ is our answer.

Video Solution (🚀Just 4 min!🚀)

https://youtu.be/P13VHc26Sgo ~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/DBHhSX8oVME

~ pi_is_3.14

See Also

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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

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