College Algebra, Chapter 9, 9.3, Section 9.3, Problem 38

Suppose the first term of a geometric sequence is $3$ and the third term is $\displaystyle \frac{4}{3}$. Find the fifth term.

Since this equation is geometric, its $n$th term is given by the formula $a_n = ar^{n-1}$. Thus,

$a_1 = ar^{1-1} = a$

$a_3 = ar^{3-1} = ar^2$

From the values we are given for these two terms, we get the following system of equations:


$
\left\{
\begin{equation}
\begin{aligned}

3 =& a
\\
\frac{4}{3} =& ar^2

\end{aligned}
\end{equation}
\right.
$



We solve this system by substituting $a = 3$ into the second equation


$
\begin{equation}
\begin{aligned}

\frac{4}{3} =& 3r^2
&& \text{Substitute } a = 3
\\
\\
\frac{4}{9} =& r^2
&& \text{Multiply both sides by } \frac{1}{3}
\\
\\
r =& \frac{2}{3}
&&

\end{aligned}
\end{equation}
$


It follows that the $n$th term of this sequence is

$\displaystyle a_n = 3 \left( \frac{2}{3} \right)^{n-1}$

Thus, the fifth term is

$\displaystyle a_5 = 3 \left( \frac{2}{3} \right)^{5-1} = 3 \left( \frac{2}{3} \right)^4 = \frac{16}{27}$