View source for 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 <math>57</math>, <math>60</math>, and <math>91</math>. What is the fourth term of this sequence?

<math>\textbf{(A) } 190 \qquad \textbf{(B) } 194 \qquad \textbf{(C) } 198 \qquad \textbf{(D) } 202 \qquad \textbf{(E) } 206</math>

==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>

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

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

* 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.

* 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.

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

~mathboy282 ~MRENTHUSIASM ~me

==Solution 2 (Answer Choices)==

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

We're looking for <math>a+3d+br^3</math>. We can substitute our value of <math>a+3d</math> in here to get <cmath>br^3-br^2-br+b+94=b(r+1)(r-1)(r-1)+94.</cmath>
Since our sequence only has positive integers we can now check by the answer choices. For each answer choice, we can subtract <math>94</math> and factor it to see if it has a perfect square factor and at least one other factor and those should differ by <math>2</math>.
<cmath>\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*}</cmath>
From this, the only possible answer choices are <math>\textbf{(A)}</math> and <math>\textbf{(E)}</math>, where <math>r=3</math>. To solve for <math>b</math>, we look back to the given equations above.

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 ==
https://youtu.be/DBHhSX8oVME

~ pi_is_3.14

== See Also ==

{{AMC10 box|year=2022|ab=A|num-b=19|num-a=21}}
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