# How do you find the 7th term of the geometric sequence with the given terms a_2 = 12, a_5 = -768?

## How do you find the 7th term of the geometric sequence with the given terms ${a}_{2} = 12 , {a}_{5} = - 768$?

Nov 23, 2015

First find the common ratio $r$...

#### Explanation:

$r = {\left[\frac{- 768}{12}\right]}^{\frac{1}{5 - 2}} = - 4$

${a}_{7} = {a}_{5} \times - 4 \times - 4 = - 768 \times 16 = - 12288$

hope that helped

Nov 23, 2015

${a}_{7} = - 12288$

#### Explanation:

Given :
${a}_{2} = 12$
${a}_{5} = - 768$

Known:
Let a constant be k
Let the geometric ratio be r

${a}_{2} \to k {r}^{2} = 12. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$ but this in not quite correct!

${a}_{5}$ is negative so $k {r}^{5} \ne + 768$

However: if we had ${a}_{n} = {\left(- 1\right)}^{n} k {r}^{n}$ then

${a}_{5} \to {\left(- 1\right)}^{5} k {r}^{5} = - 768$ would be correct

Let:
${a}_{2} = {\left(- 1\right)}^{2} k {r}^{2} = 12. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({1}_{a}\right)$
${a}_{5} = {\left(- 1\right)}^{5} k {r}^{5} = - 768. \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(2\right)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To find r}}$

color(brown)(Consider: color(white)(...)(Equation (2))/( Equation (1_a))

$\frac{{\left(- 1\right)}^{5} k {r}^{5}}{= {\left(- 1\right)}^{2} k {r}^{2}} = \frac{- 768}{12}$

$\implies - {r}^{3} = - 64$

Multiply both sides by (-1)

${r}^{3} = 64$

$\textcolor{g r e e n}{r = 4 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(4\right)}$

So $\textcolor{g r e e n}{{a}_{n} = {\left(- 1\right)}^{n} k {4}^{n} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(5\right)}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To find k}}$

$\textcolor{b r o w n}{\text{Substitute (4) into " (1_a)" giving}}$

${a}_{2} = {\left(- 1\right)}^{2} k {\left(4\right)}^{2} = 12$

$\textcolor{g r e e n}{k = \frac{12}{16} = \frac{3}{4}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(6\right)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To find } {a}_{7}}$

Substitute (4) and (6) into (5) giving

${a}_{n} = {\left(- 1\right)}^{n} k {4}^{n} \to \textcolor{b l u e}{{a}_{7} = {\left(- 1\right)}^{7} \left(\frac{3}{4}\right) {\left(4\right)}^{7}}$

${a}_{7} = \left(- 1\right) \left(\frac{3}{4}\right) \left(16384\right)$

$\textcolor{red}{{a}_{7} = - 12288}$