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:
$a_4=(\sqrt{3})^4=9$
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
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 - Page 614: 12
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
