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

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==Solution==
 
==Solution==
−
a+b=57
+
Set up a system of equations.
−
(a+n)+bm=60
+
<cmath>\begin{align*}
−
(a+2n)+bm^2=91
+
a+b&=57\\
 +
(a+n)+bm&=60\\
 +
(a+2n)+bm^2&=91
 +
\end{align*}</cmath>
  
−
b(m-1)+n=3
+
Subtract the two consecutive equations to get
−
bm(m-1)+n=31
+
<cmath>\begin{align*}
 +
b(m-1)+n&=3\\
 +
bm(m-1)+n&=31
 +
\end{align*}</cmath>
  
−
b(m-1)^2=28
+
Subtract those to get
 +
<cmath>b(m-1)^2=28</cmath>
  
−
only square with integer m that fits the factors of 28 is 4, thus b=7, (m-1)^2=4 so m=3, b=7.
+
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 \\
 +
b&=7
 +
\end{align*}</cmath>
  
−
Then, a=50 and n=3. Plug these in the second equation to get 50+n+21=60. Thus, n=-11.
+
Then, <math>a=50</math> and all the known values in the second equation to get  
  
−
so (a+3n)+bm^3=(50-33)+7*3^3=206
+
<cmath>50+n+21=60</cmath>
 +
 
 +
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*}
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Revision as of 14:39, 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*}

Video Solution by OmegaLearn

https://youtu.be/DBHhSX8oVME

~ pi_is_3.14

See Also

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

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