Difference between revisions of "2022 AMC 10A Problems/Problem 20"
Mathboy282 (talk | contribs) (→Solution) |
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Thus we conclude the answer to be | Thus we conclude the answer to be | ||
| − | \begin{align*} | + | <cmath>\begin{align*} |
(a+3n)+bm^3&=(50-33)+7 \cdot 3^3 \\ | (a+3n)+bm^3&=(50-33)+7 \cdot 3^3 \\ | ||
&= \boxed{206} | &= \boxed{206} | ||
| − | \end{align*} | + | \end{align*}</cmath> |
| + | |||
| + | +mathboy282 | ||
== 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 
, 
, and 
. What is the fourth term of this sequence?
Solution
Set up a system of equations.
Subtract the two consecutive equations to get
Subtract those to get
![\[b(m-1)^2=28\]](http://latex.artofproblemsolving.com/7/6/7/767423d3bd7f60510e0428ba9c9d495fd2162891.png)
Note that the only square with integer 
that fits the factors of 
is 
Thus, we have that
Then, 
and all the known values in the second equation to get
![\[50+n+21=60\]](http://latex.artofproblemsolving.com/8/f/7/8f715f380d31ca7c9c3809d25d50bb176b5cbdcc.png)
Thus, n=-11.
Thus we conclude the answer to be
+mathboy282
Video Solution by OmegaLearn
~ 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
