Difference between revisions of "2014 AMC 12A Problems/Problem 7"
m (→Solution) |
Sugar rush (talk | contribs) |
||
(One intermediate revision by one other user not shown) | |||
Line 2: | Line 2: | ||
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 , , 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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.