1989 AIME Problems/Problem 7

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, $\sqrt{300+k}-\sqrt{36+k}=\sqrt{596+k}-\sqrt{300+k}$ $\implies 2\sqrt{300+k} = \sqrt{596+k}+\sqrt{36+k}$ $\implies 4k+1200=596+k+36+k+2\sqrt{(596+k)(36+k)}$ $\implies 2k+568=2\sqrt{(596+k)(36+k)}$ $\implies k+284=\sqrt{(596+k)(36+k)}$ $\implies k^2+568k+80656=k^2+632k+21456$ $\implies 568k+80656=632k+21456$ $\implies 64k = 59200$ $\implies k = \boxed{925}$

Video Solution by OmegaLearn

https://youtu.be/qL0OOYZiaqA?t=251

~ pi_is_3.14

Let the arithmetic sequence be $a-d$ , $a$ , and $a+d$ . Then $(a+d)^2-a^2 = 296$ , but using the difference of squares, $d(2a+d)=296$ . Also, $a^2-(a-d)^2 = 264$ , and using the difference of squares we get $d(2a-d) = 264$ . Subtracting both equations gives $4d^2 = 32$ , $d = 4$ , and $a = 35$ . Since $a = 35$ , $a^2 = 1225 = 300+k$ and $k = \boxed{925}$ .

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

Page

Toolbox

Search

1989 AIME Problems/Problem 7

Contents

Problem

Solution 1

Solution 2 (Straighforward, but has big numbers)

Video Solution by OmegaLearn

See also