Answer :
We are given a geometric sequence with the first term
[tex]$$
a_1 = 4,
$$[/tex]
and the eighth term
[tex]$$
a_8 = -8748.
$$[/tex]
In a geometric sequence, the [tex]\( n \)[/tex]th term is given by
[tex]$$
a_n = a_1 \cdot r^{n-1},
$$[/tex]
where [tex]\( r \)[/tex] is the common ratio.
Step 1. Find the common ratio [tex]\( r \)[/tex]:
We know that
[tex]$$
a_8 = a_1 \cdot r^{7}.
$$[/tex]
Substitute the given values:
[tex]$$
-8748 = 4 \cdot r^{7}.
$$[/tex]
To solve for [tex]\( r^7 \)[/tex], divide both sides by 4:
[tex]$$
r^{7} = \frac{-8748}{4} = -2187.
$$[/tex]
Notice that [tex]\( 2187 = 3^7 \)[/tex]. Therefore, we can write
[tex]$$
r^{7} = -(3^7).
$$[/tex]
Since the seventh power of [tex]\( -3 \)[/tex] is [tex]\( (-3)^{7} = -(3^7) \)[/tex], we conclude that
[tex]$$
r = -3.
$$[/tex]
---
Step 2. Find the 16th term [tex]\( a_{16} \)[/tex]:
The formula for the [tex]\( n \)[/tex]th term is
[tex]$$
a_{16} = a_1 \cdot r^{15}.
$$[/tex]
Substitute [tex]\( a_1 = 4 \)[/tex] and [tex]\( r = -3 \)[/tex]:
[tex]$$
a_{16} = 4 \cdot (-3)^{15}.
$$[/tex]
Since an odd power of a negative number is negative, we have
[tex]$$
(-3)^{15} = -(3^{15}).
$$[/tex]
Thus,
[tex]$$
a_{16} = 4 \cdot \Big[ -\big(3^{15}\big) \Big] = -4 \cdot 3^{15}.
$$[/tex]
It is given that
[tex]$$
4 \cdot 3^{15} = 57\,395\,628.
$$[/tex]
Therefore, the 16th term is
[tex]$$
a_{16} = -57\,395\,628.
$$[/tex]
---
Final Answer:
The 16th term of the geometric sequence is
[tex]$$
\boxed{-57\,395\,628}.
$$[/tex]
[tex]$$
a_1 = 4,
$$[/tex]
and the eighth term
[tex]$$
a_8 = -8748.
$$[/tex]
In a geometric sequence, the [tex]\( n \)[/tex]th term is given by
[tex]$$
a_n = a_1 \cdot r^{n-1},
$$[/tex]
where [tex]\( r \)[/tex] is the common ratio.
Step 1. Find the common ratio [tex]\( r \)[/tex]:
We know that
[tex]$$
a_8 = a_1 \cdot r^{7}.
$$[/tex]
Substitute the given values:
[tex]$$
-8748 = 4 \cdot r^{7}.
$$[/tex]
To solve for [tex]\( r^7 \)[/tex], divide both sides by 4:
[tex]$$
r^{7} = \frac{-8748}{4} = -2187.
$$[/tex]
Notice that [tex]\( 2187 = 3^7 \)[/tex]. Therefore, we can write
[tex]$$
r^{7} = -(3^7).
$$[/tex]
Since the seventh power of [tex]\( -3 \)[/tex] is [tex]\( (-3)^{7} = -(3^7) \)[/tex], we conclude that
[tex]$$
r = -3.
$$[/tex]
---
Step 2. Find the 16th term [tex]\( a_{16} \)[/tex]:
The formula for the [tex]\( n \)[/tex]th term is
[tex]$$
a_{16} = a_1 \cdot r^{15}.
$$[/tex]
Substitute [tex]\( a_1 = 4 \)[/tex] and [tex]\( r = -3 \)[/tex]:
[tex]$$
a_{16} = 4 \cdot (-3)^{15}.
$$[/tex]
Since an odd power of a negative number is negative, we have
[tex]$$
(-3)^{15} = -(3^{15}).
$$[/tex]
Thus,
[tex]$$
a_{16} = 4 \cdot \Big[ -\big(3^{15}\big) \Big] = -4 \cdot 3^{15}.
$$[/tex]
It is given that
[tex]$$
4 \cdot 3^{15} = 57\,395\,628.
$$[/tex]
Therefore, the 16th term is
[tex]$$
a_{16} = -57\,395\,628.
$$[/tex]
---
Final Answer:
The 16th term of the geometric sequence is
[tex]$$
\boxed{-57\,395\,628}.
$$[/tex]
