Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2022 AMC 10A Problems/Problem 20

Revision as of 16:06, 5 December 2022 by Mrthinker (talk | contribs) (→‎Solution)

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

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

Solution 2

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: $br^3-br^2-br+b+94$ We can factor this to get: $b(r+1)(r-1)(r-1)+94$. 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{align*} (A) 190-94=96=2^5*3, \\ (B) 194-94=100=2^2*5^2, \\ (C) 198-94=104=2^3*13, \\ (D) 202-94=108=2^2*3^3, \\ (E) 206-94=112=2^4*7 \end{align*} From this, we know that the only possible answer choices are A and E where r=3. To solve for b, we look back to our 3 equations: \begin{align*} a+b=57, \\ a+d+3b=60, \\ a+2d+9b=91, \end{align*} We are looking for $a+3d+27b$ If 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. But,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 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_20&oldid=183637"