#### Answer

$a_n=(\sqrt{3})^{n}$
$a_4=9$

#### Work Step by Step

We are given:
$a=\sqrt{3}, r=\sqrt{3}$
We know that a geometric sequence has the form:
$a_{n}=ar^{n-1}$
Thus:
$a_n=\sqrt{3}(\sqrt{3})^{n-1}=(\sqrt{3})^{n}$
And so:
$a_4=(\sqrt{3})^4=9$

Published by
Brooks Cole

ISBN 10:
1305115546

ISBN 13:
978-1-30511-554-5

$a_n=(\sqrt{3})^{n}$
$a_4=9$

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