AIME 1989 · 第 7 题

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,

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

Solution 3

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 $2d^2 = 32$, $d = 4$, and $a = 35$. Since $a = 35$, $a^2 = 1225 = 300+k$ and $k = \boxed{925}$.

~~Disphenoid_lover

Video Solution by OmegaLearn

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

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