Question

Given the two terms in a geometric sequence how do you find the recursive formula... a1=-4 and a4=-500?

Answer 1

Recursive formula is `a_1=-4` and `a_(n+1)=5a_n`

Explanation:

In a geometric sequence, if `a` is the first term and `r` is the ratio between a term and its preceding term, then `n^(th)` term is given by `axxr^(n-1)`.

Here `a=a_1=-4` and as fourth term is `-500`

`-4xxr^(4-1)=-500` or `4r^3=500` or `r^3=125` and hence

`r=root(3)(125)=root(3)(5xx5xx5)=5`

Hence recursive formula is `a_1=-4` and `a_(n+1)=5a_n`

Answer 2

Recursive formulation:

`{ (a_1 = -4), (a_(n+1) = 5a_n color(white)(000) (n = 1,2,3,...)) :}`

Explanation:

We are given:

`{ (a_1 = -4), (a_4 = -500) :}`

The general formula for the `n`th term of a geometric series is:

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

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

A recursive formula can be given as:

`{ (a_1 = a), (a_(n+1) = ra_n color(white)(000) (n = 1,2,3,...)) :}`

In our example:

`5^3 = 125 = (-500)/(-4) = r_4/r_1 = (color(red)(cancel(color(black)(a))) r^(4-1))/(color(red)(cancel(color(black)(a))) r^(1-1)) = r^3`

So the only possible Real value for `r` is `root(3)(5^3) = 5`.

`color(white)()`
Footnote

There are two other possibilities for a geometric sequence with `a_1 = -4` and `a_4 = -500`, which are sequences of Complex numbers.

This is because `5^3` has two other cube roots, namely `5omega` and `5omega^2`, where `omega = -1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`. Either of these will also work as a suitable common ratio.