# How do you write a formula for the general term (the nth term) of the geometric sequence 4,4/3,4/27,4/81...?

Dec 5, 2015

$\frac{4}{{3}^{n - 1}}$

#### Explanation:

The formula for the general term should be of the form;

$a {r}^{n - 1}$

Where $a$ is a constant, $r$ is the rate at which the series is increasing, and $n$ is the number of the term. The constant $a$ should be the first term, in this case;

$a = 4$

We can find $r$ by dividing any term by the previous one. Lets use the first two.

r=(4"/"3)/4 color(white)(color(black)("(2nd term)")/ color(black)("(1st term)")) = 1/3

Now that we have $a$ and $r$ we can write our general formula.

$a {r}^{n - 1} = 4 {\left(\frac{1}{3}\right)}^{n - 1}$

We should double check that we get the correct series. Plugging in the first five $n$ values, we get;

$4 {\left(\frac{1}{3}\right)}^{1 - 1} , 4 {\left(\frac{1}{3}\right)}^{2 - 1} , 4 {\left(\frac{1}{3}\right)}^{3 - 1} , 4 {\left(\frac{1}{3}\right)}^{4 - 1} , 4 {\left(\frac{1}{3}\right)}^{5 - 1} \ldots$

$= \frac{4}{3} ^ 0 , \frac{4}{3} ^ 1 , \frac{4}{3} ^ 2 , \frac{4}{3} ^ 3 , \frac{4}{3} ^ 4. . .$

$= 4 , \frac{4}{3} , \frac{4}{9} , \frac{4}{27} , \frac{4}{81.} . .$

Looks like it works!