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Difference between revisions of "1989 AIME Problems/Problem 7"
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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>.
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== Video Solution by OmegaLearn ==
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https://youtu.be/qL0OOYZiaqA?t=251
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~ pi_is_3.14
  
 
== See also ==
 
== See also ==

Revision as of 06:52, 4 November 2022

Problem

If the integer $k$ is added to each of the numbers $36$, $300$, and $596$, one obtains the squares of three consecutive terms of an arithmetic series. Find $k$.

Solution

Call the terms of the arithmetic progression $a,\ a + d,\ a + 2d$, making their squares $a^2,\ a^2 + 2ad + d^2,\ a^2 + 4ad + 4d^2$.

We know that $a^2 = 36 + k$ and $(a + d)^2 = 300 + k$, and subtracting these two we get $264 = 2ad + d^2$ (1). Similarly, using $(a + d)^2 = 300 + k$ and $(a + 2d)^2 = 596 + k$, subtraction yields $296 = 2ad + 3d^2$ (2).

Subtracting the first equation from the second, we get $2d^2 = 32$, so $d = 4$. Substituting backwards yields that $a = 31$ and $k = \boxed{925}$.

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

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