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

1 Answer
Dec 18, 2016

Answer:

#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))#

#color(white)()#
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#.