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>. | + | 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}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 02:15, 28 May 2021
Problem
Suppose 
is a geometric sequence with common ratio 
and 
. If 
is an arithmetic sequence, then is
Solution
Let 
. Since are an arithmetic sequence, there is a common difference and we have 
. Dividing through by 
, we get 
or, rearranging, 
. Since we are given 
, the answer is 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
