# How do you write an nth term rule for r=8 and a_1=-2?

Aug 15, 2016

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

#### Explanation:

This is a $\textcolor{b l u e}{\text{geometric sequence}}$

with 1st term ${a}_{1} = - 2 \text{ and common ratio } r = 8$

The $\textcolor{b l u e}{\text{nth term formula}}$ is found using.

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}_{n} = a {r}^{n - 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where a is the first term and r, the common ratio.

here a = -2 and r = 8

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