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

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 $\boxed{\textbf{(B)}\ \frac{1}{3}}$

@@ Line 4: / Line 4: @@
 <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==
+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>

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

Revision as of 13:33, 15 February 2016

Problem

Solution