Question

How do you write an nth term rule for `a_1=-1/2` and `a_4=-16`?

Answer 1

`a_n = -1/2 (2root(3)(4))^(n-1)= -1/2*2^(5/3(n-1))`

Explanation:

Since this is posted under Geometric Sequences, I will assume that you are looking for a geometric sequence...

The general term of a geometric sequence is given by the formula:

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

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

In our example, we find:

`r^3 = (ar^3)/(ar^0) = a_4/a_1 = (-16)/(-1/2) = 32 = 2^3*4`

So the only Real solution is:

`r = 2root(3)(4)`

We have `a = ar^0 = a_1 = -1/2`

So the `n`th term rule can be written:

`a_n = -1/2 (2root(3)(4))^(n-1)`

or if you prefer:

`a_n = -1/2*2^(5/3(n-1))`

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Footnote

There are also two Complex solutions, corresponding to:

`r = 2omega root(3)4`

and:

`r = 2omega^2 root(3)4`

where `omega = -1/2 + sqrt(3)/2i` is the primitive Complex cube root of `1`.