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

Revision as of 07:16, 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 $r=2$ 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

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

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@@ Line 6: / Line 6: @@
 ==Solution==
-Set up a system of equations.
+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>
-<cmath>\begin{align*}
-a+b&=57\\
-(a+n)+bm&=60\\
-(a+2n)+bm^2&=91
-\end{align*}</cmath>
-Subtract the two consecutive equations to get
+We are given that
 <cmath>\begin{align*}
-b(m-1)+n&=3\\
+a+b&=57, \\
-bm(m-1)+n&=31
+a+d+br&=60, \\
+a+2d+br^2&=91,
 \end{align*}</cmath>
+and we wish to find <math>a+3d+br^3.</math>
-Subtract those to get
+Subtracting the first equation from the second and the second equation from the third, we get
-<cmath>b(m-1)^2=28</cmath>
-Note that the only square with integer <math>m</math> that fits the factors of <math>28</math> is <math>4.</math> Thus, we have that
 <cmath>\begin{align*}
-m&=3 \\
+d+b(r-1)&=3, \\
-b&=7
+d+br(r-1)&=31.
 \end{align*}</cmath>
+Subtract these results, we get <cmath>b(r-1)^2=28.</cmath>
-Then, <math>a=50</math> and all the known values in the second equation to get
+Note that <math>r=2</math> or <math>b=3.</math> We proceed with casework:
-<cmath>50+n+21=60</cmath>
-Thus, <math>n=-11.</math>
+* 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.
-Thus we conclude the answer to be
+* 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> The answer is <math>a+3d+br^3=17+189=\boxed{\textbf{(E) } 206}.</math>
-<cmath>\begin{align*}
-(a+3n)+bm^3&=(50-33)+7 \cdot 3^3 \\
-&= \boxed{206}
-\end{align*}</cmath>
-+mathboy282
+~mathboy282 ~MRENTHUSIASM
 == Video Solution by OmegaLearn ==

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

Revision as of 07:16, 19 November 2022

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