Difference between revisions of "2022 AMC 10A Problems/Problem 20"
Mathboy282 (talk | contribs) (→Video Solution by OmegaLearn) |
Mathboy282 (talk | contribs) (→Solution) |
||
| Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
| − | + | a+b=57 | |
| + | (a+n)+bm=60 | ||
| + | (a+2n)+bm^2=91 | ||
| + | |||
| + | b(m-1)+n=3 | ||
| + | bm(m-1)+n=31 | ||
| + | |||
| + | b(m-1)^2=28 | ||
| + | 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. Then, a=50 and n=3 so (a+3n)+bm^3=(50+3)+7*3^3= | ||
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == | ||
Revision as of 14:34, 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
a+b=57 (a+n)+bm=60 (a+2n)+bm^2=91
b(m-1)+n=3 bm(m-1)+n=31
b(m-1)^2=28 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. Then, a=50 and n=3 so (a+3n)+bm^3=(50+3)+7*3^3=
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. 
