We know the formula for a geometric sequence is
an = a1 * r ^ (n-1)
We know the 4th term is 234
234 = a1 * r^ (4-1) = a1 * r^3
We know the 6th term is 108
108 = a1 * r^ (6-1) = a1 (6-1) = a1 * r^5
Divide the 2nd equation by the 1st equation
108 = a1 * r^5
----------------------
234 = a1 * r^3
(108/234) = r^5/ r^3
(6/13) = r^2
Taking the square root of each side
±sqrt(6/13) = r
The decimal approximation is
.67936 to the nearest tenth is .7
Now we need to find a1
234 = a1 * r^3
Using + sqrt(6/13)
234 = a1 ( 6/13) ^(3/2)
Using a decimal approximation to the nearest tenth
a1 = 746.3
Using - sqrt(6/13)
234 = a1 (-( 6/13)) ^(3/2)
Using a decimal approximation to the nearest tenth
a1 = -746.3
We will get two possible equations
an = 746.3 ( sqrt(6/13))^ (n-1)
or
an = -746.3 ( -sqrt(6/13))^ (n-1)