# How do you find a formula of the nth term if the 4th term in the geometric sequence is -192 and the 9th term is 196608?

Apr 22, 2016

$\left\{3 , - 12 , 48 , - 192 , \ldots . , 196608 , \ldots . , 3 \times {\left(- 4\right)}^{n - 1} , \ldots\right\}$

#### Explanation:

Let the nth term be $a {r}^{n - 1}$

$a {r}^{3} = - 192 \mathmr{and} a {r}^{8} = 196608$.

Dividing, ${r}^{5} = - 1024 \to r = - 4 \mathmr{and} a = 3$.

May 1, 2016

${a}_{n} = 3 {\left(- 4\right)}^{n - 1}$ or four other Complex possibilities.

#### Explanation:

The general term of a geometric sequence can be described by the formula:

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

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

Given:

$\left\{\begin{matrix}{a}_{4} = - 192 \\ {a}_{9} = 196608\end{matrix}\right.$

We find:

${r}^{5} = \frac{a {r}^{8}}{a {r}^{3}} = {a}_{9} / {a}_{4} = \frac{196608}{- 192} = - 1024 = {\left(- 4\right)}^{5}$

This yields one possible Real value for the common ratio:

$r = \sqrt{{\left(- 4\right)}^{5}} = - 4$

Then:

$a = \frac{a {r}^{3}}{{r}^{3}} = {a}_{4} / {\left(- 4\right)}^{3} = \frac{- 192}{- 64} = 3$

So if our sequence is of Real numbers, then the only solution is:

${a}_{n} = 3 {\left(- 4\right)}^{n - 1}$

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Complex soutions

There are $4$ other possible solutions, corresponding to the $4$ other fifth roots of $- 1024$.

Let $\xi = \frac{\sqrt{5} - 1}{4} + \frac{\sqrt{10 + 2 \sqrt{5}}}{4} i = \cos \left(\frac{2 \pi}{5}\right) + i \sin \left(\frac{2 \pi}{5}\right)$

This is a primitive Complex $5$th root of $1$

Then the possible values for the common ratio are:

$- 4 {\xi}^{k}$ with $k = 0 , 1 , 2 , 3 , 4$

with corresponding initial terms:

$3 {\xi}^{- 3 k}$

The case $k = 0$ gives us the Real solution.

Since ${\xi}^{5} = 1$, the possible formulas for the general term can be written::

$\left\{\begin{matrix}{a}_{n} = 3 {\left(- 4\right)}^{n - 1} \\ {a}_{n} = 3 {\xi}^{2} {\left(- 4 \xi\right)}^{n - 1} \\ {a}_{n} = 3 {\xi}^{4} {\left(- 4 {\xi}^{2}\right)}^{n - 1} \\ {a}_{n} = 3 \xi {\left(- 4 {\xi}^{3}\right)}^{n - 1} \\ {a}_{n} = 3 {\xi}^{3} {\left(- 4 {\xi}^{4}\right)}^{n - 1}\end{matrix}\right.$