# How do you write an nth term rule for a_1=-1/2 and a_4=-16?

Dec 18, 2016

${a}_{n} = - \frac{1}{2} {\left(2 \sqrt{4}\right)}^{n - 1} = - \frac{1}{2} \cdot {2}^{\frac{5}{3} \left(n - 1\right)}$

#### Explanation:

Since this is posted under Geometric Sequences, I will assume that you are looking for 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.

In our example, we find:

${r}^{3} = \frac{a {r}^{3}}{a {r}^{0}} = {a}_{4} / {a}_{1} = \frac{- 16}{- \frac{1}{2}} = 32 = {2}^{3} \cdot 4$

So the only Real solution is:

$r = 2 \sqrt{4}$

We have $a = a {r}^{0} = {a}_{1} = - \frac{1}{2}$

So the $n$th term rule can be written:

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

or if you prefer:

${a}_{n} = - \frac{1}{2} \cdot {2}^{\frac{5}{3} \left(n - 1\right)}$

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Footnote

There are also two Complex solutions, corresponding to:

$r = 2 \omega \sqrt{4}$

and:

$r = 2 {\omega}^{2} \sqrt{4}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.