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

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