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

Revision as of 13:40, 12 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

Set up a system of equations. $\begin{align*} a+b&=57\\ (a+n)+bm&=60\\ (a+2n)+bm^2&=91 \end{align*}$

Subtract the two consecutive equations to get $\begin{align*} b(m-1)+n&=3\\ bm(m-1)+n&=31 \end{align*}$

Subtract those to get $b(m-1)^2=28$

Note that the only square with integer $m$ that fits the factors of $28$ is $4.$ Thus, we have that $\begin{align*} m&=3 \\ b&=7 \end{align*}$

Then, $a=50$ and all the known values in the second equation to get

$50+n+21=60$

Thus, $n=-11.$

Thus we conclude the answer to be $\begin{align*} (a+3n)+bm^3&=(50-33)+7 \cdot 3^3 \\ &= \boxed{206} \end{align*}$

+mathboy282

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 32: / Line 32: @@
 <cmath>50+n+21=60</cmath>
-Thus, n=-11.
+Thus, <math>n=-11.</math>
 Thus we conclude the answer to be

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

Revision as of 13:40, 12 November 2022

Contents

Problem

Solution

Video Solution by OmegaLearn

See Also