# What is the 12th term of a geometric sequence where a_1 = 8 and a_6 = -8,192?

Oct 29, 2015

Negative 33554432

#### Explanation:

Let K be a constant.
Let the geometric ratio be r.
Let ${a}_{n}$ be the resulting ${n}^{t h} \textcolor{w h i t e}{x}$term of the sequence

Then at

$n = 1$ we have ${a}_{n} = k {r}^{n} = \left(+ 8\right)$
$n = 6$ we have ${a}_{n} = k {r}^{n} = \left(- 8192\right)$

The fact that ${a}_{6}$ is negative implies a switching between positive and negative sequence. This is achieved by the incorporation of ${\left(- 1\right)}^{n + 1}$. It has to be $n + 1$ or an appropriate variation to force ${a}_{n}$ to be positive and negative in the appropriate place.

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To find r

Known:
${a}_{1} = {\left(- 1\right)}^{1 + 1} \left(k\right) \left({r}^{1}\right) = \left(+ 8\right) \text{ "-> " } k r = 8$ ........(1)
${a}_{6} = {\left(- 1\right)}^{6 + 1} \left(k\right) \left({r}^{6}\right) = - 8192 \text{ -> " } - k {r}^{6} = - 8192$..(2)

To isolate r use $\left(2\right) \div i \mathrm{de} \left(1\right)$ giving:

$\frac{- k {r}^{6}}{k r} = \frac{- 8192}{8}$

Thus ${r}^{5} = 1025$ Note the reversal from negative to positive.

so $r = \sqrt[5]{1025} = 4$

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To find k by substitution into (1)
${\left(- 1\right)}^{2} \left(k\right) \left(4\right) = 8$
k=2

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Thus:
${a}_{12} = {\left(- 1\right)}^{12 + 1} \left(2\right) {\left(4\right)}^{12} = - 33554432$

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