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

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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 <math>r=2</math> or <math>b=3.</math> We proceed with casework:
+
Note that either <math>b=28</math> or <math>b=7.</math> We proceed with casework:
  
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* If <math>r=2,</math> then <math>b=28,a=29,</math> and <math>d=25.</math> The arithmetic sequence is <math>29,4,-21,-46,</math> arriving at a contradiction.
+
* 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.
  
−
* If <math>r=3,</math> then <math>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.
+
* 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>

Revision as of 08:35, 19 November 2022

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

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 $b=28$ or $b=7.$ We proceed with casework:

  • If $b=28,$ then $r=2,a=29,$ and $d=25.$ The arithmetic sequence is $29,4,-21,-46,$ arriving at a contradiction.
  • If $b=7,$ then $r=3,a=50,$ 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

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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