What process do I use to find the missing numbers in a geometric sequence, when there are not two consecutive numbers? For example:81, __, __, __,...

In geometric progression we know
that:

an = a1* r^(n-1)

such
that (an) is any term in the progression, a1 is the first term, and r is the common
difference.

In the first
example:

Given a1=
81

==> a5 = 1

Then we
will need to find the common difference r , then we will calculate the other
terms.

==> a5 = a1*
r^4

==> 1=
81*r^4

==> r^4 =
1/81

==> r = +-1/3

Now
since we have 2 values for r, then we have two possible
sequences.

==> When r=
1/3

==> a1=
81

==> a2= 81*1/3  =
27

==> a3= 81*1/3^2 =
9

==> a4= 81*1/3^3 =
3

==> a5 = 81^1/3^4 =
1

==> The sequence is: 81, 27, 9, 3, 1
with a common difference r= 1/3

Now when r=
-1/3

==> a1=
81

==> a2= 81*-1/3 =
-27

==> a3= 81*(-1/3)^2 =
9

==> a4= 81*(-1/3)^3 =
-3

==> a5= 81*(-1/3)^4 =
1

Then the sequence is: 81, -27, 9, -3, 1
with the common difference r=
-1/3

 

Now
for the example number 2:

a1=
9/4

a4= 2/3

We will determine
r.

==> a4=
a1*r^3

==> 2/3 = 9/4 *
r^3

==> r^3 =
8/27

==> r= 2/3

Then
the common difference is 2/3

==> a1=
9/4

==> a2= 9/4* 2/3 =
3/2

==> a3= 9/4 * (2/3)^2 =
1

==> a4= 9/4* (2/3)^3 =
2/3

==> a5= 9/4*(2/3)^4 =
4/9

Then the sequence is: 9/4, 3/2, 1, 2/3,
4/9  which common difference r= 2/3