1994 AHSME Problems/Problem 20

Problem

Suppose $x,y,z$ is a geometric sequence with common ratio $r$ and $x \neq y$ . If $x, 2y, 3z$ is an arithmetic sequence, then $r$ is

$\textbf{(A)}\ \frac{1}{4} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{1}{2} \qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ 4$

Solution

Let $y=xr, z=xr^2$ . Since $x, 2y, 3z$ are an arithmetic sequence, there is a common difference and we have $2xr-x=3xr^2-2xr$ . Dividing through by $x$ , we get $2r-1=3r^2-2r$ or, rearranging, $(r-1)(3r-1)=0$ . Since we are given $x\neq y\implies r\neq 1$ , the answer is $\boxed{\textbf{(B)}\ \frac{1}{3}}$

1994 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1994 AHSME Problems/Problem 20

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