To find the appropriate formula for the [tex]$n$[/tex]th term of a geometric sequence where the fifth term is given as [tex]\(\frac{1}{4}\)[/tex], we need to consider the standard formula for the [tex]$n$[/tex]th term of a geometric sequence:
[tex]\[a_n = a \cdot r^{(n-1)}\][/tex]
where:
- [tex]\(a\)[/tex] is the first term,
- [tex]\(r\)[/tex] is the common ratio,
- [tex]\(n\)[/tex] is the term number.
Given that the fifth term ([tex]\(a_5\)[/tex]) is [tex]\(\frac{1}{4}\)[/tex], we can write:
[tex]\[a_5 = a \cdot r^4 = \frac{1}{4}\][/tex]
Now, let's evaluate each of the given formulas by substituting [tex]\(n = 5\)[/tex] to find out which one satisfies the given condition [tex]\(a_5 = \frac{1}{4}\)[/tex].
1. First formula: [tex]\(a_n = 18\left(\frac{1}{4}\right)^{n-1}\)[/tex]
[tex]\[a_5 = 18\left(\frac{1}{4}\right)^{5-1} = 18\left(\frac{1}{4}\right)^4 = 18\left(\frac{1}{256}\right) = \frac{18}{256} \approx 0.0703125\][/tex]
Clearly, [tex]\(\frac{18}{256}\)[/tex] does not equal [tex]\(\frac{1}{4}\)[/tex], so this formula is incorrect.
2. Second formula: [tex]\(a_n = \frac{1}{16}\left(\frac{1}{4}\right)^{n-1}\)[/tex]
[tex]\[a_5 = \frac{1}{16}\left(\frac{1}{4}\right)^{5-1} = \frac{1}{16}\left(\frac{1}{4}\right)^4 = \frac{1}{16} \cdot \frac{1}{256} = \frac{1}{4096} \approx 0.000244140625\][/tex]
This also does not equal [tex]\(\frac{1}{4}\)[/tex], so this formula is incorrect.
3. Third formula: [tex]\(a_n = \frac{1}{4} \cdot 16^{n-1}\)[/tex]
[tex]\[a_5 = \frac{1}{4} \cdot 16^{5-1} = \frac{1}{4} \cdot 16^4 = \frac{1}{4} \cdot 65536 = 16384.0\][/tex]
This is way off from [tex]\(\frac{1}{4}\)[/tex], so this formula is incorrect.
4. Fourth formula: [tex]\(a_n = \frac{1}{4} \cdot \left(\frac{1}{16}\right)^{n-1}\)[/tex]
[tex]\[a_5 = \frac{1}{4} \cdot \left(\frac{1}{16}\right)^{5-1} = \frac{1}{4} \cdot \left(\frac{1}{16}\right)^4 = \frac{1}{4} \cdot \frac{1}{65536} = \frac{1}{262144} \approx 0.000003814697265625\][/tex]
This does not equal [tex]\(\frac{1}{4}\)[/tex], so this formula is also incorrect.
So, none of these formulas directly meets the condition of giving [tex]\(\frac{1}{4}\)[/tex] as the fifth term according to the calculations provided.
