# 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?

##### 2 Answers

#### Answer:

#### Explanation:

Let the nth term be

Dividing,

#### Answer:

#### Explanation:

The general term of a geometric sequence can be described by the formula:

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

where

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)#

**Complex soutions**

There are

Let

This is a primitive Complex

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

Since

#{ (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)) :}#