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

The general formula for a geometric sequence is `a_n = a_1 * r^(n-1)`, where `a_n` is the `n^(th)` 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 * r^(3-1)`
`a_3 = a_1 * r^2`
`color(blue)(24 = a_1 * r^2)`

`a_5 = a_1 * r^(5-1)`
`a_5 = a_1 * r^4`
`color(blue)(96= a_1 * r^4)`

Now we can solve the system of equations:

`color(blue)(24 = a_1 * r^2)` `->` solve for `a_1`

`a_1=24/r^2`

`color(blue)(96= a_1 * r^4)`

`96=24/r^2 * r^4` `->` substitute the value of `a_1` into the second equation

`96=24 * r^2`

`4=r^2`

`r=+-2`

Now that we have the value of `r`, we can find the value of `a_1`. Using the first equation, `color(blue)(24 = a_1 * r^2)`, we get

`24 = a_1 * r^2`

`24 = a_1 * (+-2)^2`

`24 = a_1 * 4`

`a_1=6`

So our formula for the sequence can be either `color(red)(a_n = 6 * 2^(n-1))` or `color(red)(a_n = 6 * (-2)^(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.

`color(red)(a_n = 6 * 2^(n-1))`

The common ratio is `2`, so start with `6` and multiply each term by `2 => 6, 12, 24, 48, 96`

`color(red)(a_n = 6 * (-2)^(n-1))`

The common ratio is `-2`, so start with `6` and multiply each term by `-2 => 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 = 48/24 = 2`, or the terms are `24, -48, 96`, meaning that `r=-48/24 = -2`.

After you find `r`, you can find `a_1` the same way we did above. In the end, you get `color(red)(a_n = 6 * 2^(n-1))` or `color(red)(a_n = 6 * (-2)^(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.)