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

(Solution)
 
(2 intermediate revisions by 2 users 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
  
== Solution ==
+
==See Also==
so the terms are
+
{{AMC12 box|year=2014|ab=A|num-b=6|num-a=8}}
 
+
{{MAA Notice}}
<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>
 

Latest revision as of 23:16, 25 November 2020

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

The terms are $\sqrt 3$, $\sqrt[3]3$, and $\sqrt[6]3$, which are equivalent to $3^{\frac{3}{6}}$, $3^{\frac{2}{6}}$, and $3^{\frac{1}{6}}$. So the next term will be $3^{\frac{0}{6}}=1$, so the answer is $\boxed{\textbf{(A)}}$.

Video Solution

https://youtu.be/rJytKoJzNBY

See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png