Difference between revisions of "2014 AMC 12A Problems/Problem 7"
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The first three terms of a geometric progression are <math>\sqrt 3</math>, <math>\sqrt[3]3</math>, and <math>\sqrt[6]3</math>. What is the fourth term? | The first three terms of a geometric progression are <math>\sqrt 3</math>, <math>\sqrt[3]3</math>, and <math>\sqrt[6]3</math>. What is the fourth term? | ||
| − | <math>\textbf{(A) }1\qquad | + | <math>\textbf{(A) }1\qquad\textbf{(B) }\sqrt[7]3\qquad\textbf{(C) }\sqrt[8]3\qquad\textbf{(D) }\sqrt[9]3\qquad\textbf{(E) }\sqrt[10]3\qquad</math> |
| − | \textbf{(B) }\sqrt[7]3\qquad | ||
| − | \textbf{(C) }\sqrt[8]3\qquad | ||
| − | \textbf{(D) }\sqrt[9]3\qquad | ||
| − | \textbf{(E) }\sqrt[10]3\qquad</math> | ||
| + | ==Solution== | ||
| + | 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>. | ||
| + | ==Video Solution== | ||
| + | https://youtu.be/rJytKoJzNBY | ||
| − | == | + | ==See Also== |
| − | + | {{AMC12 box|year=2014|ab=A|num-b=6|num-a=8}} | |
| − | + | {{MAA Notice}} | |
| − | |||
| − | |||
Latest revision as of 23:16, 25 November 2020
Contents
Problem 7
The first three terms of a geometric progression are 
, ![$\sqrt[3]3$](http://latex.artofproblemsolving.com/9/d/f/9df2b093dad7aff038bf8e76013debc0f33465dc.png)
, and ![$\sqrt[6]3$](http://latex.artofproblemsolving.com/f/7/6/f76bf02aeb1d529aef2a4953c053ddaef97334fb.png)
. What is the fourth term?
![$\textbf{(A) }1\qquad\textbf{(B) }\sqrt[7]3\qquad\textbf{(C) }\sqrt[8]3\qquad\textbf{(D) }\sqrt[9]3\qquad\textbf{(E) }\sqrt[10]3\qquad$](http://latex.artofproblemsolving.com/1/c/e/1ce555f7402c58e7f2bd569b30fd28c2495d9532.png)
Solution
The terms are , , and , which are equivalent to 
, 
, and 
. So the next term will be 
, so the answer is 
.
Video Solution
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
