Difference between revisions of "1987 AHSME Problems/Problem 9"
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\textbf{(D)}\ \frac{2}{3}\qquad | \textbf{(D)}\ \frac{2}{3}\qquad | ||
\textbf{(E)}\ 2</math> | \textbf{(E)}\ 2</math> | ||
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| + | == Solution == | ||
| + | To get from the second term to the fourth term, we add <math>2</math> lots of the common difference, so if the common difference is <math>d</math>, we have <math>2d = 2x -x = x \implies d = \frac{x}{2}</math>. Thus the sequence is <math>\frac{x}{2}, x, \frac{3x}{2}, 2x</math>, so the ratio is <math>\frac{\frac{x}{2}}{\frac{3x}{2}} = \frac{1}{3}</math>, which is answer <math>\boxed{\text{B}}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 13:06, 1 March 2018
Problem
The first four terms of an arithmetic sequence are 
. The ratio of 
to 
is
Solution
To get from the second term to the fourth term, we add 
lots of the common difference, so if the common difference is 
, we have 
. Thus the sequence is 
, so the ratio is 
, which is answer 
.
See also
| 1987 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
