Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 12 - Section 12.3 - Geometric Sequences - 12.3 Exercises - Page 864: 12

Answer

a. $a_{n} = \sqrt 3(\sqrt 3)^{n-1}$ b. Fourth term: 9

Work Step by Step

a. $a_{n} = a(r)^{n-1}$ is the formula for the nth term of a geometric sequence. In this case a = $\sqrt{3}$ and r=$\sqrt{3}$. Plug these values into the formula to get a formula for the nth term of this sequence. b. To find the fourth term plug in n = 4 into the sequence: $a_{n} = \sqrt{3}(\sqrt{3})^{4-1} = \sqrt{3}(\sqrt{3})^{3} = 9$
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