How do you write an equation for the nth term of geometric sequence -6561, 1458, -324?

Mar 8, 2016

$\textcolor{b l u e}{{\left(- 1\right)}^{n} a {r}^{n - 1}}$

Explanation:

The general structure of a geometric sequence is:

$a {r}^{0} \text{ , "ar^1" , "ar^2" , "ar^3".....} a {r}^{n} .$

So lets just look at the numbers for the moment

From the above it is quite evident that you can find $r$ by applying:

$\text{ } \frac{a {r}^{1}}{a {r}^{0}} = r$

So we have

$\text{Term (n) a " r^(n-1)" } a {r}^{n - 1}$

$\text{ 1 -6561 0 -6561}$

$\text{ 2 -6561 r +1458}$

$\text{ 3 -6561 "r^2" -324}$

Ignoring the signs to start with. We can allow for that later.
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$\text{ } \frac{a {r}^{1}}{a {r}^{0}} = \frac{1458}{6561} = \frac{2}{9} = r$

Test: The next number should be of magnitude 324

$\text{ } 6561 \times {\left(\frac{2}{9}\right)}^{2} = 324$ as required
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Consider the alternating positive negative

If we use ${\left(- 1\right)}^{n}$ this will give us the correct sign

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$\textcolor{b l u e}{\text{Final structure}}$

$\textcolor{b l u e}{{\left(- 1\right)}^{n} a {r}^{n - 1}}$