Difference between revisions of "1989 AIME Problems/Problem 7"
Pi is 3.14 (talk | contribs) (→Solution) |
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Subtracting the first equation from the second, we get <math>2d^2 = 32</math>, so <math>d = 4</math>. Substituting backwards yields that <math>a = 31</math> and <math>k = \boxed{925}</math>. | Subtracting the first equation from the second, we get <math>2d^2 = 32</math>, so <math>d = 4</math>. Substituting backwards yields that <math>a = 31</math> and <math>k = \boxed{925}</math>. | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/qL0OOYZiaqA?t=251 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
== See also == | == See also == | ||
Revision as of 06:52, 4 November 2022
Problem
If the integer 
is added to each of the numbers 
, 
, and 
, one obtains the squares of three consecutive terms of an arithmetic series. Find .
Solution
Call the terms of the arithmetic progression 
, making their squares 
.
We know that 
and 
, and subtracting these two we get 
(1). Similarly, using and 
, subtraction yields 
(2).
Subtracting the first equation from the second, we get 
, so 
. Substituting backwards yields that 
and 
.
Video Solution by OmegaLearn
https://youtu.be/qL0OOYZiaqA?t=251
~ pi_is_3.14
See also
| 1989 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
