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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>
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==Solution==
 
==Solution==
 
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>
 
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==
 
==See Also==
  
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{{AHSME box|year=1994|num-b=16|num-a=18}}
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{{AHSME box|year=1994|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:44, 6 March 2025

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 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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