How do you find a formula of the nth term if the 4th term in the geometric sequence is -192 and the 9th term is 196608?

Let the nth term be `a r^(n-1)`

`a r^3=-192 and a r^8=196608`.

Dividing, `r^5=-1024 to r=-4 and a=3`.

The general term of a geometric sequence can be described by the formula:

`a_n = a r^(n-1)`

where `a` is the initial term and `r` the common ratio.

Given:

`{ (a_4 = -192), (a_9 = 196608) :}`

We find:

`r^5 = (a r^8)/(a r^3) = a_9/a_4 = 196608/(-192) = -1024 = (-4)^5`

This yields one possible Real value for the common ratio:

`r = root(5)((-4)^5) = -4`

Then:

`a = (a r^3) / (r^3) = a_4/(-4)^3 = (-192)/(-64) = 3`

So if our sequence is of Real numbers, then the only solution is:

`a_n = 3(-4)^(n-1)`

`color(white)()`
Complex soutions

There are `4` other possible solutions, corresponding to the `4` other fifth roots of `-1024`.

Let `xi = (sqrt(5)-1)/4 + sqrt(10+2sqrt(5))/4 i = cos((2pi)/5) + i sin((2pi)/5)`

This is a primitive Complex `5`th root of `1`

Then the possible values for the common ratio are:

`-4xi^k` with `k=0,1,2,3,4`

with corresponding initial terms:

`3xi^(-3k)`

The case `k=0` gives us the Real solution.

Since `xi^5 = 1`, the possible formulas for the general term can be written::

`{ (a_n = 3(-4)^(n-1)), (a_n = 3xi^2(-4xi)^(n-1)), (a_n = 3xi^4(-4xi^2)^(n-1)), (a_n = 3xi(-4xi^3)^(n-1)), (a_n = 3xi^3(-4xi^4)^(n-1)) :}`