2022 AMC 10A Problems/Problem 20

Revision as of 13:40, 12 November 2022 by Mathboy282 (talk | contribs) (Solution)

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

See Also

2022 AMC 10A (ProblemsAnswer KeyResources)
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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