Sure, let's analyze each given sequence and determine the formula for the nth term based on the patterns observed:
### Sequence 1: [tex]\(3, 4, 5, 6, 7, \ldots\)[/tex]
This is an arithmetic sequence where each term increases by 1. The first term ([tex]\(a\)[/tex]) is 3, and the common difference ([tex]\(d\)[/tex]) is 1.
The formula for the nth term of an arithmetic sequence is given by:
[tex]\[ a_n = a + (n - 1) \cdot d \][/tex]
For this sequence:
- [tex]\(a = 3\)[/tex]
- [tex]\(d = 1\)[/tex]
So,
[tex]\[ a_n = 3 + (n - 1) \cdot 1 = 3 + n - 1 = n + 2 \][/tex]
Thus, the nth term for Sequence 1 is:
[tex]\[ a_n = n + 2 \][/tex]
### Sequence 2: [tex]\(3, 5, 7, 9, 11, \ldots\)[/tex]
This is another arithmetic sequence where each term increases by 2. The first term ([tex]\(a\)[/tex]) is 3, and the common difference ([tex]\(d\)[/tex]) is 2.
The formula for the nth term of an arithmetic sequence is given by:
[tex]\[ a_n = a + (n - 1) \cdot d \][/tex]
For this sequence:
- [tex]\(a = 3\)[/tex]
- [tex]\(d = 2\)[/tex]
So,
[tex]\[ a_n = 3 + (n - 1) \cdot 2 = 3 + 2n - 2 = 2n + 1 \][/tex]
Thus, the nth term for Sequence 2 is:
[tex]\[ a_n = 2n + 1 \][/tex]
### Sequence 3: [tex]\(2, 4, 8, 16, 31, \ldots\)[/tex]
This appears to be a geometric sequence initially, but the 5th term doesn't fit that pattern. The first four terms are doubling each time:
[tex]\[ 2, 4, 8, 16 \][/tex]
However, the 5th term is 31. To generalize this sequence, it's given that:
- The nth term for the first four terms follows the pattern [tex]\(2^n\)[/tex].
- From the 5th term onward, the sequence diverges from the geometric pattern and we have to use an adjusted formula.
The 5th term is derived such that:
[tex]\[ a_5 = 17 \][/tex]
Thus, the nth term for Sequence 3 is as follows:
[tex]\[ a_n = 2^n \text{ for } n \leq 4 \][/tex]
[tex]\[ a_n \text{ changes to a different pattern for } n > 4 \text{ yielding } a_5 = 17 \][/tex]
### Sequence 4: [tex]\(-1, 1, -1, 1, -1, \ldots\)[/tex]
This sequence alternates between -1 and 1.
If [tex]\( n \)[/tex] is odd, the term is -1.
If [tex]\( n \)[/tex] is even, the term is 1.
Thus, the nth term for Sequence 4 is:
[tex]\[ a_n = (-1)^{n+1} \][/tex]
### Sequence 5: [tex]\(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots\)[/tex]
This sequence is a harmonic sequence where the nth term is the reciprocal of [tex]\( n \)[/tex].
Thus, the nth term for Sequence 5 is:
[tex]\[ a_n = \frac{1}{n} \][/tex]
