# How do you write a rule for the nth term of the geometric term given the two terms a_3=25, a_6=-25/64?

Feb 24, 2018

$400 {\left(- \frac{1}{4}\right)}^{n - 1}$

#### Explanation:

The nth term of a geometric series is given by:

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

Where $\boldsymbol{a}$ is the first term, $\boldsymbol{r}$ is the common ratio and $\boldsymbol{n}$ is the nth term.

We have:

Third term $= 25$

Sixth term $= - \frac{25}{64}$

$\therefore$

${a}_{3} = a {r}^{2} = 25 \textcolor{w h i t e}{888888.8} \left[1\right]$

${a}_{6} = a {r}^{5} = - \frac{25}{64} \textcolor{w h i t e}{8888} \left[2\right]$

Solving for $\boldsymbol{a}$ and $\boldsymbol{r}$

$\left[1\right]$

$a = \frac{25}{r} ^ 2$

Substituting in $\left[2\right]$

$\left(\frac{25}{r} ^ 2\right) {r}^{5} = - \frac{25}{64}$

Multiply by ${r}^{2} / 25$

${r}^{5} = - {r}^{2} / 64$

$64 {r}^{5} + {r}^{2} = 0$

Factor out $\boldsymbol{r}$

$r \left(64 {r}^{4} + r\right) = 0$

We know $\boldsymbol{r}$ can't be zero.

$\left(64 {r}^{4} + r\right) = 0$

Factor out $\boldsymbol{r}$ again.

$r \left({64}^{3} + 1\right) = 0$

As before:

$64 {r}^{3} + 1 = 0$

${r}^{3} = - \frac{1}{64}$

$r = \sqrt[3]{- \frac{1}{64}} = - \frac{1}{4}$

Using this in $\left[1\right]$

$a {\left(- \frac{1}{4}\right)}^{2} = 25$

$a = \frac{25}{- \frac{1}{4}} ^ 2 = \frac{25}{\frac{1}{64}} = 400$

So we have:

$400 {\left(- \frac{1}{4}\right)}^{n - 1}$ for our nth term.