Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 12 - Review - Exercises: 27

Answer

$a_{5}=\displaystyle \frac{81}{4}$

Work Step by Step

(see p. 858) A geometric sequence is a sequence whose terms are obtained by multiplying each term by the same fixed constant $r$ to get the next term. A geometric sequence has the form $a, ar, ar^{2}, ar^{3}, \ldots$ The number $a$ is the first term of the sequence, and the number $r $is the common ratio. The nth term of the sequence is $\quad a_{n}=ar^{n-1}$ ---------- Given $a_{3}=9$, and $r=\displaystyle \frac{3}{2},$ From $ a_{3}=ar^{3-1}$, we find $a,$ $9=a(\displaystyle \frac{3}{2})^{2}$ $9=a\displaystyle \cdot\frac{9}{4}\qquad.../\times\frac{4}{9}$ $4=a$ From $a_{n}=ar^{n-1}$, we find $a_{5}$ $a_{5}=4\displaystyle \times(\frac{3}{2})^{4}=4\times\frac{81}{16}=\frac{81}{4}$
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