# How do you find nth term rule for a_1=1/4 and a_3=6?

Aug 2, 2016

The formula for the $n$th term is one of the following:

${a}_{n} = \frac{1}{4} {\left(2 \sqrt{6}\right)}^{n - 1}$

${a}_{n} = \frac{1}{4} {\left(- 2 \sqrt{6}\right)}^{n - 1}$

#### Explanation:

Assuming this is a geometric sequence...

The general term of a geometric sequence is given by the formula:

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

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

So we have $a = {a}_{1} = \frac{1}{4}$ and we find:

${r}^{2} = \frac{a {r}^{2}}{a {r}^{0}} = {a}_{3} / {a}_{1} = \frac{6}{\frac{1}{4}} = 24$

So:

$r = \pm \sqrt{24} = \pm 2 \sqrt{6}$

So the formula for the $n$th term is one of the following:

${a}_{n} = \frac{1}{4} {\left(2 \sqrt{6}\right)}^{n - 1}$

${a}_{n} = \frac{1}{4} {\left(- 2 \sqrt{6}\right)}^{n - 1}$