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)) :}#