Difference between revisions of "2014 AMC 12A Problems/Problem 7"
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The terms are <math>\sqrt 3</math>, <math>\sqrt[3]3</math>, and <math>\sqrt[6]3</math>, which are equivalent to <math>3^{\frac{3}{6}}</math>, <math>3^{\frac{2}{6}}</math>, and <math>3^{\frac{1}{6}}</math>. So the next term will be <math>3^{\frac{0}{6}}=1</math>, so the answer is <math>\boxed{\textbf{(A)}}</math>. | The terms are <math>\sqrt 3</math>, <math>\sqrt[3]3</math>, and <math>\sqrt[6]3</math>, which are equivalent to <math>3^{\frac{3}{6}}</math>, <math>3^{\frac{2}{6}}</math>, and <math>3^{\frac{1}{6}}</math>. So the next term will be <math>3^{\frac{0}{6}}=1</math>, so the answer is <math>\boxed{\textbf{(A)}}</math>. | ||
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+ | ==See Also== | ||
+ | {{AMC12 box|year=2014|ab=A|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Revision as of 13:29, 8 February 2014
Problem 7
The first three terms of a geometric progression are , , and . What is the fourth term?
Solution
The terms are , , and , which are equivalent to , , and . So the next term will be , so the answer is .
See Also
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 | |
All AMC 12 Problems and Solutions |
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