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

2014 AMC 12A Problem 7

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The first three terms of a geometric progression are

\sqrt{3},

\sqrt[3]{3},

and

\sqrt[6]{3}.

What is the fourth term?

1

\sqrt[7]{3}

\sqrt[8]{3}

\sqrt[9]{3}

\sqrt[10]{3}

Rewrite each term as a power of

3

3^{\frac{1}{2}},

3^{\frac{1}{3}},

3^{\frac{1}{6}}

The common ratio is

3^{\frac{1}{3}-\frac{1}{2}}=3^{-\frac{1}{6}}

; multiply the third term by it

Writing the terms as powers of

3,

they are

3^{\frac{1}{2}},

3^{\frac{1}{3}},

3^{\frac{1}{6}}.

The common ratio is

\dfrac{3^{\frac{1}{3}}}{3^{\frac{1}{2}}}=3^{-\frac{1}{6}}.

The fourth term is

3^{\frac{1}{6}}\cdot3^{-\frac{1}{6}}=3^{0}=1.

Thus, the correct answer is A.