# How do you write an nth term rule for r=1/8 and a_1=4?

Oct 17, 2017

${a}_{n} = 4 {\left(\frac{1}{8}\right)}^{n - 1}$

#### Explanation:

presumably this is a geometric sequence question

in which case
for a GP

${a}_{1} =$the first term#

$r =$ the common ratio

we ahve

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

in this case we re given

${a}_{1} = 4$

$r = \frac{1}{8}$

so ${a}_{n} = 4 {\left(\frac{1}{8}\right)}^{n - 1}$