Difference between revisions of "1989 AIME Problems/Problem 7"
Pi is 3.14 (talk | contribs) (→Solution) |
|||
Line 8: | Line 8: | ||
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 07:52, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.