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1978 AHSME Problems/Problem 24

Revision as of 19:38, 11 June 2018 by Apple2017 (talk | contribs) (Created page with "Let the geometric progression be <math>a,</math> <math>ar,</math> <math>ar^2,</math> so <math>a = x(y - z)</math>, <math>ar = y(z - x)</math>, and <math>ar^2 = z(x - y).</math...")
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Let the geometric progression be $a,$ $ar,$ $ar^2,$ so $a = x(y - z)$, $ar = y(z - x)$, and $ar^2 = z(x - y).$ Adding these equations, we get \[a + ar + ar^2 = xy - xz + yz - xy + xz - yz = 0.\] Since $a$ is nonzero, we can divide by $a$, to get $\boxed{r^2 + r + 1 = 0}$. The answer is (A).

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