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Which formula can be used to find the [tex]$n$[/tex]th term of a geometric sequence where the first term is 8 and the common ratio is -3?

A. [tex]a_n = 8 \cdot (-3)^{n-1}[/tex]
B. [tex]a_n = -3 \cdot (8)^{n-1}[/tex]
C. [tex]a_n = (8 \cdot (-3))^{n-1}[/tex]
D. [tex]a_n = 8 \cdot (-3)^n[/tex]

Sagot :

First, let's review what we know about geometric sequences. A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio.

The formula to find the [tex]\( n \)[/tex]th term ([tex]\( a_n \)[/tex]) of a geometric sequence is:

[tex]\[ a_n = a_1 \cdot r^{(n-1)} \][/tex]

Here:
- [tex]\( a_1 \)[/tex] is the first term of the sequence.
- [tex]\( r \)[/tex] is the common ratio.
- [tex]\( n \)[/tex] is the term number we want to find.

Given information:
- The first term ([tex]\( a_1 \)[/tex]) is 8.
- The common ratio ([tex]\( r \)[/tex]) is -3.

Substituting the given values into the formula, we get:

[tex]\[ a_n = 8 \cdot (-3)^{(n-1)} \][/tex]

Now, let's examine the choices provided:

1. [tex]\( a_n = 8 \cdot (-3)^{n-1} \)[/tex]
2. [tex]\( a_n = -3 \cdot (8)^{n-1} \)[/tex]
3. [tex]\( a_n = (8 \cdot (-3))^{n-1} \)[/tex]
4. [tex]\( a_n = 8 \cdot (-3)^n \)[/tex]

Comparing these options to our derived formula:

- Option 1: [tex]\( a_n = 8 \cdot (-3)^{n-1} \)[/tex]
- This matches exactly with our derived formula.
- Option 2: [tex]\( a_n = -3 \cdot (8)^{n-1} \)[/tex]
- This does not match because the base of the exponent should be the common ratio, which is -3, not 8.
- Option 3: [tex]\( a_n = (8 \cdot (-3))^{n-1} \)[/tex]
- This does not match because it suggests multiplying the first term by the common ratio and then raising it to the power of [tex]\( n-1 \)[/tex].
- Option 4: [tex]\( a_n = 8 \cdot (-3)^n \)[/tex]
- This does not match because the exponent is [tex]\( n \)[/tex] instead of [tex]\( n-1 \)[/tex].

Therefore, the correct formula to find the [tex]\( n \)[/tex]th term of the geometric sequence is:

[tex]\[ a_n = 8 \cdot (-3)^{n-1} \][/tex]

Thus, the answer is option 1: [tex]\( a_n = 8 \cdot (-3)^{n-1} \)[/tex].