Difference between revisions of "1994 AHSME Problems/Problem 20"

(Created page with "==Problem== Suppose <math>x,y,z</math> is a geometric sequence with common ratio <math>r</math> and <math>x \neq y</math>. If <math>x, 2y, 3z</math> is an arithmetic sequence, th...")
 
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<math> \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</math>
 
<math> \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</math>
 
==Solution==
 
==Solution==
 +
Let <math>y=xr, z=xr^2</math>. Then we have <math>2xr-x=3xr^2-2xr</math>. Dividing through by <math>x</math>, we get <math>2r-1=3r^2-2r, 3r^2-4r+1=0, 3(r-1)(r-\frac{1}{3})</math>. Since we are given <math>x\neq y\implies r\neq 1</math>, the answer is <math>\boxed{\textbf{(B)}\ \frac{1}{3}}</math>

Revision as of 13:33, 15 February 2016

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 $y=xr, z=xr^2$. Then we have $2xr-x=3xr^2-2xr$. Dividing through by $x$, we get $2r-1=3r^2-2r, 3r^2-4r+1=0, 3(r-1)(r-\frac{1}{3})$. Since we are given $x\neq y\implies r\neq 1$, the answer is $\boxed{\textbf{(B)}\ \frac{1}{3}}$

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