## College Algebra 7th Edition

$a_n=(\sqrt{3})^{n}$ $a_4=9$
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: $a_4=(\sqrt{3})^4=9$