# How do you write a rule for the nth term of the geometric sequence, then find a6 given 5, -5/3, 5/9, -5/27, ...?

Jun 6, 2017

${a}_{n} = 5 {\left(- \frac{1}{3}\right)}^{n - 1} \text{ and } {a}_{6} = - \frac{5}{243}$

#### Explanation:

$\text{for the standard geometric sequence}$

$a , a r , a {r}^{2} , \ldots \ldots , a {r}^{n - 1} \leftarrow \text{ nth term}$

$\text{where " a=a_1" and r is the common ratio}$

$r = \frac{{a}_{2}}{{a}_{1}} = \frac{{a}_{3}}{{a}_{2}} = \ldots . = \frac{{a}_{n}}{{a}_{n + 1}}$

$\Rightarrow r = \frac{- \frac{5}{3}}{5} = - \frac{1}{3}$

$\Rightarrow {a}_{n} = a {r}^{n - 1} = 5 {\left(- \frac{1}{3}\right)}^{5} \leftarrow \text{ nth term rule}$

$\Rightarrow {a}_{6} = 5 \times - \frac{1}{243} = - \frac{5}{243}$