Difference between revisions of "2021 AIME I Problems/Problem 5"
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==Solution== | ==Solution== | ||
| − | Let the terms be <math>a-b</math>, <math>a</math>, and <math>a+b</math>. Then we want <math>(a-b)^2+a^2+(a+b)^2=ab^2</math>, or <math>3a^2+2b^2=ab^2</math>. Rearranging, we get <math> | + | Let the terms be <math>a-b</math>, <math>a</math>, and <math>a+b</math>. Then we want <math>(a-b)^2+a^2+(a+b)^2=ab^2</math>, or <math>3a^2+2b^2=ab^2</math>. Rearranging, we get <math>b^2=\frac{3a^2}{a-2}</math>. Simplifying further, <math>b^2=3a+6+\frac{12}{a-2}</math>. |
==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=4|num-a=6}} | {{AIME box|year=2021|n=I|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:33, 11 March 2021
Problem
Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.
Solution
Let the terms be 
, 
, and 
. Then we want 
, or 
. Rearranging, we get 
. Simplifying further, 
.
See also
| 2021 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
