# How do you write a rule for the nth term of the geometric term given the two terms a_3=24, a_5=96?

Aug 5, 2017

${a}_{n} = 6 \cdot {2}^{n - 1}$ or ${a}_{n} = 6 \cdot {\left(- 2\right)}^{n - 1}$

#### Explanation:

The general formula for a geometric sequence is ${a}_{n} = {a}_{1} \cdot {r}^{n - 1}$, where ${a}_{n}$ is the ${n}^{t h}$ term, ${a}_{1}$ is the first term, and $r$ is the common ratio.

I'm going to explain how to do this problem two ways.

The Long Way

Since we are given ${a}_{3} = 24$ and ${a}_{5} = 96$, we can substitute them into the formula.

${a}_{3} = {a}_{1} \cdot {r}^{3 - 1}$
${a}_{3} = {a}_{1} \cdot {r}^{2}$
$\textcolor{b l u e}{24 = {a}_{1} \cdot {r}^{2}}$

${a}_{5} = {a}_{1} \cdot {r}^{5 - 1}$
${a}_{5} = {a}_{1} \cdot {r}^{4}$
$\textcolor{b l u e}{96 = {a}_{1} \cdot {r}^{4}}$

Now we can solve the system of equations:

$\textcolor{b l u e}{24 = {a}_{1} \cdot {r}^{2}}$ $\to$ solve for ${a}_{1}$

${a}_{1} = \frac{24}{r} ^ 2$

$\textcolor{b l u e}{96 = {a}_{1} \cdot {r}^{4}}$

$96 = \frac{24}{r} ^ 2 \cdot {r}^{4}$ $\to$ substitute the value of ${a}_{1}$ into the second equation

$96 = 24 \cdot {r}^{2}$

$4 = {r}^{2}$

$r = \pm 2$

Now that we have the value of $r$, we can find the value of ${a}_{1}$. Using the first equation, $\textcolor{b l u e}{24 = {a}_{1} \cdot {r}^{2}}$, we get

$24 = {a}_{1} \cdot {r}^{2}$

$24 = {a}_{1} \cdot {\left(\pm 2\right)}^{2}$

$24 = {a}_{1} \cdot 4$

${a}_{1} = 6$

So our formula for the sequence can be either $\textcolor{red}{{a}_{n} = 6 \cdot {2}^{n - 1}}$ or $\textcolor{red}{{a}_{n} = 6 \cdot {\left(- 2\right)}^{n - 1}}$.

To verify if these are correct, you can write out the first few terms and see if they match the information given in the problem.

$\textcolor{red}{{a}_{n} = 6 \cdot {2}^{n - 1}}$

The common ratio is $2$, so start with $6$ and multiply each term by $2 \implies 6 , 12 , 24 , 48 , 96$

$\textcolor{red}{{a}_{n} = 6 \cdot {\left(- 2\right)}^{n - 1}}$

The common ratio is $- 2$, so start with $6$ and multiply each term by $- 2 \implies 6 , - 12 , 24 , - 48 , 96$

In both of these formulas, ${a}_{3} = 24$ and ${a}_{5} = 96$.

The Short Way

We are given ${a}_{3}$ and ${a}_{5}$, so we can easily find out ${a}_{4}$ in order to get the value of $r$.

${a}_{3} , {a}_{4} , {a}_{5}$

$24 , {a}_{4} , 96$

To find ${a}_{4}$, we can simply calculate the geometric mean.

(a_4)^2 = 24 * 96 => a_4 = +-sqrt(24 * 96) = +-sqrt2304 = +-48

So the three terms are either $24 , 48 , 96$, meaning that $r = \frac{48}{24} = 2$, or the terms are $24 , - 48 , 96$, meaning that $r = - \frac{48}{24} = - 2$.

After you find $r$, you can find ${a}_{1}$ the same way we did above. In the end, you get $\textcolor{red}{{a}_{n} = 6 \cdot {2}^{n - 1}}$ or $\textcolor{red}{{a}_{n} = 6 \cdot {\left(- 2\right)}^{n - 1}}$.

(This method is easier in the context of this problem, but if you were given terms such as ${a}_{10}$ and ${a}_{19}$, you would definitely want to use the first method.)