2014 AMC 12A Problems/Problem 7

Revision as of 00:16, 26 November 2020 by Sugar rush (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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