College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.3 - Geometric Sequences - 8.3 Exercises: 12

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