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

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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>
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Let <math>y=xr, z=xr^2</math>. Since <math>x, 2y, 3z</math> are an arithmetic sequence, there is a common difference and we have <math>2xr-x=3xr^2-2xr</math>. Dividing through by <math>x</math>, we get <math>2r-1=3r^2-2r</math> or, rearranging, <math>(r-1)(3r-1)=0</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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==See Also==
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{{AHSME box|year=1994|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 03:15, 28 May 2021

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


See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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