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

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 $k = \boxed{925}$ .

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
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1989 AIME Problems/Problem 7

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