2022 AMC 10A Problems/Problem 20

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 $r=2$ or $b=3.$ We proceed with casework:

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

If $r=3,$ then $b=7,a=50,$ and $d=-11.$ The arithmetic sequence is $50,39,28,17,$ and the geometric sequence is $7,21,63,189.$ 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

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2022 AMC 10A Problems/Problem 20

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