# 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

~ pi_is_3.14

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