# How do you find the Nth term in the geometric sequence where the first term is 3 and the fourth term is #6sqrt2#?

##### 1 Answer

Dec 31, 2015

#### Explanation:

The general term of a geometric sequence can be written:

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

where

In our case we are given

So we find:

#r^3 = (ar^3)/(ar^0) = a_4/a_1 = (6sqrt(2))/3 = 2sqrt(2) = (sqrt(2))^3#

So (assuming the geometric sequence is Real):

#r = root(3)(sqrt(2)^3) = sqrt(2)#

and

#a_n = 3 * (sqrt(2))^(n-1) = 3 * 2^((n-1)/2)#