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Difference between revisions of "1989 AIME Problems/Problem 7"

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If the integer <math>k</math> is added to each of the numbers <math>36</math>, <math>300</math>, and <math>596</math>, one obtains the squares of three consecutive terms of an arithmetic series. Find <math>k</math>.
 
If the integer <math>k</math> is added to each of the numbers <math>36</math>, <math>300</math>, and <math>596</math>, one obtains the squares of three consecutive terms of an arithmetic series. Find <math>k</math>.
  
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== Solution ==
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== Solution 1==
 
Call the terms of the [[arithmetic progression]] <math>a,\ a + d,\ a + 2d</math>, making their squares <math>a^2,\ a^2 + 2ad + d^2,\ a^2 + 4ad + 4d^2</math>.
 
Call the terms of the [[arithmetic progression]] <math>a,\ a + d,\ a + 2d</math>, making their squares <math>a^2,\ a^2 + 2ad + d^2,\ a^2 + 4ad + 4d^2</math>.
  
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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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==Solution 2 (Straighforward, but has big numbers)==
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Since terms in an arithmetic progression have constant differences,
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<cmath>\sqrt{300+k}-\sqrt{36+k}=\sqrt{596+k}-\sqrt{300+k}</cmath>
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<cmath>\implies 2\sqrt{300+k} = \sqrt{596+k}+\sqrt{36+k}</cmath>
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<cmath>\implies 4k+1200=596+k+36+k+2\sqrt{(596+k)(36+k)}</cmath>
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<cmath>\implies 2k+568=2\sqrt{(596+k)(36+k)}</cmath>
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<cmath>\implies k+284=\sqrt{(596+k)(36+k)}</cmath>
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<cmath>\implies k^2+568k+80656=k^2+632k+21456</cmath>
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<cmath>\implies 568k+80656=632k+21456</cmath>
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<cmath>\implies 64k = 59200</cmath>
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<cmath>\implies k = 925</cmath>
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Revision as of 15:09, 2 June 2023

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 1

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 $k = \boxed{925}$.

Solution 2 (Straighforward, but has big numbers)

Since terms in an arithmetic progression have constant differences, \[\sqrt{300+k}-\sqrt{36+k}=\sqrt{596+k}-\sqrt{300+k}\] \[\implies 2\sqrt{300+k} = \sqrt{596+k}+\sqrt{36+k}\] \[\implies 4k+1200=596+k+36+k+2\sqrt{(596+k)(36+k)}\] \[\implies 2k+568=2\sqrt{(596+k)(36+k)}\] \[\implies k+284=\sqrt{(596+k)(36+k)}\] \[\implies k^2+568k+80656=k^2+632k+21456\] \[\implies 568k+80656=632k+21456\] \[\implies 64k = 59200\] \[\implies k = 925\]

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
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All AIME Problems and Solutions

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