Difference between revisions of "2014 AMC 12A Problems/Problem 7"

(Solution)
Line 14: Line 14:
  
 
<math>\sqrt 3</math>, <math>\sqrt[3]3</math>, and <math>\sqrt[6]3</math> which is equivalent to <math>3^{\frac{3}{6}}</math>, <math>3^{\frac{2}{6}}</math>, and <math>3^{\frac{1}{6}}</math>
 
<math>\sqrt 3</math>, <math>\sqrt[3]3</math>, and <math>\sqrt[6]3</math> which is 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}}</math> which is equal to <math>1\textbf{(A) }\qquad</math>
+
so the next term will be <math>3^{\frac{0}{6}}</math> which is equal to <math>1\textbf{(A) }\qquad</math>

Revision as of 17:47, 7 February 2014

Problem 7

The first three terms of a geometric progression are $\sqrt 3$, $\sqrt[3]3$, and $\sqrt[6]3$. What is the fourth term?

$\textbf{(A) }1\qquad \textbf{(B) }\sqrt[7]3\qquad \textbf{(C) }\sqrt[8]3\qquad \textbf{(D) }\sqrt[9]3\qquad \textbf{(E) }\sqrt[10]3\qquad$


Solution

so the terms are

$\sqrt 3$, $\sqrt[3]3$, and $\sqrt[6]3$ which is equivalent to $3^{\frac{3}{6}}$, $3^{\frac{2}{6}}$, and $3^{\frac{1}{6}}$ so the next term will be $3^{\frac{0}{6}}$ which is equal to $1\textbf{(A) }\qquad$