# How do you find nth term rule for a_2=5 and a_4=1/5?

Aug 16, 2016

There are two possibilities:

${a}_{n} = 25 {\left(\frac{1}{5}\right)}^{n - 1}$

${a}_{n} = - 25 {\left(- \frac{1}{5}\right)}^{n - 1}$

#### Explanation:

Assuming that this is a geomtric sequence, since that's the topic under which it is posted...

The general formula for the $n$th term of a geometric sequence is:

${a}_{n} = a {r}^{n - 1}$

where $a$ is the initial term and $r$ the common ratio.

In our example, we find:

${r}^{2} = \frac{a {r}^{3}}{a {r}^{1}} = {a}_{4} / {a}_{2} = \frac{\frac{1}{5}}{5} = \frac{1}{25}$

Hence $r = \pm \sqrt{\frac{1}{25}} = \pm \frac{1}{5}$

If $r = \frac{1}{5}$ then $a = \frac{a r}{r} = {a}_{2} / r = \frac{5}{\frac{1}{5}} = 25$

If $r = - \frac{1}{5}$ then $a = \frac{a r}{r} = {a}_{2} / r = \frac{5}{- \frac{1}{5}} = - 25$