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

Latest revision as of 22:29, 6 October 2023

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 (Algebra)

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 (Answer Choices)

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 $br^3-br^2-br+b+94=b(r+1)(r-1)(r-1)+94.$ 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 $2$ . $\begin{alignat*}{8} \textbf{(A)} \ 190-94&=96&&=2^5\cdot3, \\ \textbf{(B)} \ 194-94&=100&&=2^2\cdot5^2, \\ \textbf{(C)} \ 198-94&=104&&=2^3\cdot13, \\ \textbf{(D)} \ 202-94&=108&&=2^2\cdot3^3, \\ \textbf{(E)} \ 206-94&=112&&=2^4\cdot7. \end{alignat*}$ From this, the only possible answer choices are $\textbf{(A)}$ and $\textbf{(E)}$ , where $r=3$ . To solve for $b$ , we look back to the given equations above.

We are looking for $a+3d+27b$ . If $\textbf{(A)}$ 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 $3$ . However, 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 (🚀Just 4 min!🚀)

https://youtu.be/P13VHc26Sgo ~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/DBHhSX8oVME

~ pi_is_3.14

@@ Line 5: / Line 5: @@
 <math>\textbf{(A) } 190 \qquad \textbf{(B) } 194 \qquad \textbf{(C) } 198 \qquad \textbf{(D) } 202 \qquad \textbf{(E) } 206</math>
-==Solution==
+==Solution 1 (Algebra)==
-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>
+We are given that
 <cmath>\begin{align*}
-a+b&=57\\
+a+b&=57, \\
-(a+n)+bm&=60\\
+a+d+br&=60, \\
-(a+2n)+bm^2&=91
+a+2d+br^2&=91,
 \end{align*}</cmath>
+and we wish to find <math>a+3d+br^3.</math>
-Subtract the two consecutive equations to get
+Subtracting the first equation from the second and the second equation from the third, we get
 <cmath>\begin{align*}
-b(m-1)+n&=3\\
+d+b(r-1)&=3, \\
-bm(m-1)+n&=31
+d+br(r-1)&=31.
 \end{align*}</cmath>
+Subtract these results, we get <cmath>b(r-1)^2=28.</cmath>
-Subtract those to get
+Note that either <math>b=28</math> or <math>b=7.</math> We proceed with casework:
-<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
+* 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.
-<cmath>\begin{align*}
-m&=3 \\
+* 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.
-b&=7
-\end{align*}</cmath>
-Then, <math>a=50</math> and all the known values in the second equation to get
+Therefore, The answer is <math>a+3d+br^3=17+189=\boxed{\textbf{(E) } 206}.</math>
-<cmath>50+n+21=60</cmath>
+~mathboy282 ~MRENTHUSIASM
-Thus, n=-11.
+==Solution 2 (Answer Choices)==
-Thus we conclude the answer to be
+Start similarly to Solution 1 and deduce the three equations
 <cmath>\begin{align*}
-(a+3n)+bm^3&=(50-33)+7 \cdot 3^3 \\
+a+b&=57, \\
-&= \boxed{206}
+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.
-+mathboy282
+==Video Solution (🚀Just 4 min!🚀)==
+https://youtu.be/P13VHc26Sgo
+~<i>Education, the Study of Everything </i>
 == Video Solution by OmegaLearn ==

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