# How do you find the first term of a geometric sequence whose fourth term is -6 and whose common ratio is 1/3?

Nov 20, 2015

The first term is $\left(- 162\right)$

#### Explanation:

Suppose the first term is ${a}_{1}$
then (since the common ratio is $\frac{1}{3}$)

$\textcolor{w h i t e}{\text{XXX}} {a}_{2} = \left(\frac{1}{3}\right) \times {a}_{1}$

$\textcolor{w h i t e}{\text{XXX}} {a}_{3} = \left(\frac{1}{3}\right) \times {a}_{2} = {\left(\frac{1}{3}\right)}^{2} \times {a}_{1}$

$\textcolor{w h i t e}{\text{XXX}} {a}_{4} = \left(\frac{1}{3}\right) \times {a}_{3} = {\left(\frac{1}{3}\right)}^{3} \times {a}_{1}$

We are told
$\textcolor{w h i t e}{\text{XXX}} {a}_{4} = - 6$

So
$\textcolor{w h i t e}{\text{XXX}} {\left(\frac{1}{3}\right)}^{3} \times {a}_{1} = - 6$

$\textcolor{w h i t e}{\text{XXX}} \frac{1}{27} \times {a}_{1} = - 6$

$\textcolor{w h i t e}{\text{XXX}} {a}_{1} = \left(- 6\right) \times 27 = - 162$