How do you write an nth term rule for a_1=-1/2a1=−12 and a_4=-16a4=−16?

1 Answer
George C.
Dec 18, 2016

a_n = -1/2 (2root(3)(4))^(n-1)= -1/2*2^(5/3(n-1))an=−12(23√4)n−1=−12⋅253(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)an=arn−1

where aa is the initial term and rr the common ratio.

In our example, we find:

r^3 = (ar^3)/(ar^0) = a_4/a_1 = (-16)/(-1/2) = 32 = 2^3*4r3=ar3ar0=a4a1=−16−12=32=23⋅4

So the only Real solution is:

r = 2root(3)(4)r=23√4

We have a = ar^0 = a_1 = -1/2a=ar0=a1=−12

So the nnth term rule can be written:

a_n = -1/2 (2root(3)(4))^(n-1)an=−12(23√4)n−1

or if you prefer:

a_n = -1/2*2^(5/3(n-1))an=−12⋅253(n−1)

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Footnote

There are also two Complex solutions, corresponding to:

r = 2omega root(3)4r=2ω3√4

and:

r = 2omega^2 root(3)4r=2ω23√4

where omega = -1/2 + sqrt(3)/2iω=−12+√32i is the primitive Complex cube root of 11.

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