# How do you write a rule for the nth term of the geometric sequence and then find a_5 given a_4=-3, r=-2?

Sep 10, 2017

${a}_{n} = \frac{3}{8} {\left(- 2\right)}^{n - 1} , {a}_{5} = 6$

#### Explanation:

$\text{the terms in a standard geometric sequence are}$

$a , a r , a {r}^{2} , a {r}^{3} , \ldots \ldots , a {r}^{n - 1}$

$\text{where a is the first term and r the common ratio}$

$\text{the nth term is }$

•color(white)(x)a_n=ar^(n-1)

${a}_{4} = a {r}^{3} = - 3 \Rightarrow a = \frac{- 3}{- 2} ^ 3 = \frac{- 3}{- 8} = \frac{3}{8}$

$\Rightarrow {a}_{n} = \frac{3}{8} {\left(- 2\right)}^{n - 1}$

$\Rightarrow {a}_{5} = \frac{3}{8} {\left(- 2\right)}^{4} = \frac{3}{8} \times 16 = 6$