College Algebra, Chapter 9, 9.3, Section 9.3, Problem 40

Suppose the common ratio in a geometric sequence is $\displaystyle \frac{3}{2}$ and the fifth term is $1$. Find the first three terms.

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

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

$\displaystyle 1 = a \left( \frac{3}{2} \right)^4$

Solve for the first term $a$


$\left\{ \begin{equation} \begin{aligned} 1 =& a \left( \frac{81}{16} \right) \\ \frac{16}{81} =& a \qquad \text{Multiply both sides by } \frac{16}{81} \end{aligned} \end{equation} \right.$


For the second term,


$\begin{equation} \begin{aligned} a_2 =& \frac{16}{81} \left( \frac{3}{2} \right)^{2-1} \\ \\ =& \frac{8}{27} \end{aligned} \end{equation}$


For the third term,


$\begin{equation} \begin{aligned} a_3 =& \frac{16}{81} \left( \frac{3}{2} \right)^{3-1} \\ \\ =& \frac{16}{81} \left( \frac{9}{4} \right) \\ \\ =& \frac{4}{9} \end{aligned} \end{equation}$


So the first three terms of the geometric sequence,

$\displaystyle \frac{16}{81}, \frac{8}{27}, \frac{4}{9}$