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

Revision as of 20:52, 5 December 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 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: $br^3-br^2-br+b+94=b(r+1)(r-1)(r-1)+94$ . 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 by OmegaLearn

https://youtu.be/DBHhSX8oVME

~ pi_is_3.14

@@ Line 41: / Line 41: @@
 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>
+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: <math>br^3-br^2-br+b+94=b(r+1)(r-1)(r-1)+94</math>. 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>.
+We're looking for <math>a+3d+br^3</math>. We can substitute our value of <math>a+3d</math> in here to get: <math>br^3-br^2-br+b+94=b(r+1)(r-1)(r-1)+94</math>. 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, \\
@@ Line 52: / Line 53: @@
 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 A 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.
+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 by OmegaLearn ==

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

Page

Toolbox

Search