Let's address each of the geometric series one by one and find out how many terms are in each series.
### Part (a)
We are given the geometric series: [tex]\( 4, 8, 16, \ldots, 1024 \)[/tex]
- First term ([tex]\(a\)[/tex]): 4
- Common ratio ([tex]\(r\)[/tex]): [tex]\(\frac{8}{4} = 2\)[/tex]
- Last term: 1024
In a geometric series, the [tex]\(n\)[/tex]-th term ([tex]\(T_n\)[/tex]) is given by:
[tex]\[ T_n = a \cdot r^{n-1} \][/tex]
We need to find [tex]\(n\)[/tex] such that:
[tex]\[ 1024 = 4 \cdot 2^{n-1} \][/tex]
Dividing both sides by 4:
[tex]\[ 256 = 2^{n-1} \][/tex]
Since [tex]\(256 = 2^8\)[/tex], we have:
[tex]\[ 2^{n-1} = 2^8 \][/tex]
Therefore:
[tex]\[ n - 1 = 8 \][/tex]
[tex]\[ n = 9 \][/tex]
So, the number of terms in the series is 9.
### Part (b)
We are given the geometric series: [tex]\( 81, 54, 36, \ldots, \frac{512}{243} \)[/tex]
- First term ([tex]\(a\)[/tex]): 81
- Common ratio ([tex]\(r\)[/tex]): [tex]\(\frac{54}{81} = \frac{2}{3}\)[/tex]
- Last term: [tex]\(\frac{512}{243}\)[/tex]
In a geometric series, the [tex]\(n\)[/tex]-th term ([tex]\(T_n\)[/tex]) is given by:
[tex]\[ T_n = a \cdot r^{n-1} \][/tex]
We need to find [tex]\(n\)[/tex] such that:
[tex]\[ \frac{512}{243} = 81 \cdot \left(\frac{2}{3}\right)^{n-1} \][/tex]
Dividing both sides by 81:
[tex]\[ \left(\frac{2}{3}\right)^{n-1} = \frac{512}{243 \cdot 81} \][/tex]
Calculating the right-hand side:
[tex]\[ 243 \cdot 81 = 19683 \][/tex]
[tex]\[ \left(\frac{512}{19683}\right) = \left(\frac{2}{3}\right)^9 \][/tex]
Therefore:
[tex]\[ \left(\frac{2}{3}\right)^{n-1} = \left(\frac{2}{3}\right)^9 \][/tex]
This implies:
[tex]\[ n - 1 = 9 \][/tex]
[tex]\[ n = 10 \][/tex]
So, the number of terms in the series is 10.
### Part (c)
The series has only one given term: [tex]\( 729 \)[/tex]
Since there is only one term provided with no indication of continuation or other data, it is impossible to determine the number of terms in the series.
So, the number of terms cannot be determined with the given information.
