Given:
The first term
�
1
=
729
a
1
=729
The seventh term
�
7
=
1
a
7
=1
We need to find the common ratio
�
r and the intermediate terms
�
2
,
�
3
,
�
4
,
�
5
,
a
2
,a
3
,a
4
,a
5
, and
�
6
a
6
.
The general formula for the
�
n-th term of a geometric sequence is:
�
�
=
�
1
⋅
�
(
�
−
1
)
a
n
=a
1
⋅r
(n−1)
For
�
7
=
1
a
7
=1:
�
7
=
�
1
⋅
�
6
a
7
=a
1
⋅r
6
1
=
729
⋅
�
6
1=729⋅r
6
�
6
=
1
729
r
6
=
729
1
�
=
(
1
729
)
1
6
r=(
729
1
)
6
1
First, we need to calculate
�
r:
�
=
(
1
729
)
1
6
=
72
9
−
1
6
r=(
729
1
)
6
1
=729
−
6
1
Now, let's compute
�
r:
729
=
3
6
729=3
6
72
9
−
1
6
=
(
3
6
)
−
1
6
=
3
−
1
=
1
3
729
−
6
1
=(3
6
)
−
6
1
=3
−1
=
3
1
So,
�
=
1
3
r=
3
1
.
Now, we can find the intermediate terms:
�
2
=
�
1
⋅
�
=
729
⋅
1
3
=
243
a
2
=a
1
⋅r=729⋅
3
1
=243
�
3
=
�
1
⋅
�
2
=
729
⋅
(
1
3
)
2
=
729
⋅
1
9
=
81
a
3
=a
1
⋅r
2
=729⋅(
3
1
)
2
=729⋅
9
1
=81
�
4
=
�
1
⋅
�
3
=
729
⋅
(
1
3
)
3
=
729
⋅
1
27
=
27
a
4
=a
1
⋅r
3
=729⋅(
3
1
)
3
=729⋅
27
1
=27
�
5
=
�
1
⋅
�
4
=
729
⋅
(
1
3
)
4
=
729
⋅
1
81
=
9
a
5
=a
1
⋅r
4
=729⋅(
3
1
)
4
=729⋅
81
1
=9
�
6
=
�
1
⋅
�
5
=
729
⋅
(
1
3
)
5
=
729
⋅
1
243
=
3
a
6
=a
1
⋅r
5
=729⋅(
3
1
)
5
=729⋅
243
1
=3
Thus, the 5 geometric means between 729 and 1 are:
243
,
81
,
27
,
9
,
3
243,81,27,9,3
Step-by-step explanation: