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 $\quad a_{n}=ar^{n-1}$
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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 $a_{5}$
$a_{5}=4\displaystyle \times(\frac{3}{2})^{4}=4\times\frac{81}{16}=\frac{81}{4}$
