The sum of the first 5 terms of a geometric sequence is 93 and the 10th term is 32. What is the common ratio and what is the fourth term?
Answers
If
3
2
is a typo for
3
32
then the common ratio is
1
2
and the
4
th term is
6
.
Explanation:
The general term of a geometric sequence is expressible by the formula:
a
n
=
a
r
n
−
1
where
a
is the initial term and
r
the common ratio.
If
r
≠
1
then the sum of the first
N
terms is given by the formula:
N
∑
n
=
1
a
n
=
N
∑
n
=
1
a
r
n
−
1
=
a
r
N
−
1
r
−
1
In our example, we are given:
93
=
5
∑
n
=
1
a
n
=
a
r
5
−
1
r
−
1
a
r
9
=
a
10
=
3
2
If this is to have rational solutions, then we should look at the factors of
93
.
The prime factorisation of
93
is:
93
=
3
⋅
31
=
3
⋅
2
5
−
1
2
−
1
So
93
is the sum of the first five terms of the geometric sequence:
3
,
6
,
12
,
24
,
48
But if that were our sequence, then
a
10
=
32
9
=
1536
- somewhat larger than
3
2
.
How about if we reverse the sequence?
48
,
24
,
12
,
6
,
3
Then the
6
th term will be
3
2
, not the
10
th term - which will be
48
⋅
(
1
2
)
9
=
3
32
Is there a typo in the question? Should the
10
th term be specified as
3
32
?
If so, then the common ratio is
1
2
and the
4
th term is
6
.
If the question is correct in the form given then the answer will be much more messy. Specifically, the common ratio
r
would be the Real zero of the nonic polynomial:
62
r
9
−
r
4
−
r
3
−
r
2
−
r
−
1
, an irrational number approximately
0.709461
.