# How do you find a_9 when a_1 = 1000, r = 1/3?

May 25, 2016

${a}_{9} = \frac{1000}{6561}$

#### Explanation:

For the standard geometric sequence.

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

where a is 1st term

r ,the common ratio $= {a}_{2} / {a}_{1} = {a}_{3} / {a}_{2} = \ldots . = {a}_{n} / {a}_{n - 1}$

and the nth term ${a}_{n} = a {r}^{n - 1}$

here n = 9 , a = 1000 and $r = \frac{1}{3}$

$\Rightarrow {a}_{9} = 1000 {\left(\frac{1}{3}\right)}^{8} = \frac{1000}{3} ^ 8 = \frac{1000}{6561}$