Sure! Let's go through the process of finding the [tex]\( n^{\text{th}} \)[/tex] term for each of the given sequences.
### Part (a): Sequence [tex]\(3, 6, 9, 12, \ldots\)[/tex]
This sequence is an arithmetic sequence where the first term [tex]\( a \)[/tex] is 3 and the common difference [tex]\( d \)[/tex] is 3.
The general formula for the [tex]\( n^{\text{th}} \)[/tex] term of an arithmetic sequence is:
[tex]\[ a_n = a + (n-1) \cdot d \][/tex]
For this sequence:
[tex]\[ a_n = 3 + (n-1) \cdot 3 \][/tex]
Therefore, the [tex]\( n^{\text{th}} \)[/tex] term is:
[tex]\[ a_n = 3 + 3n - 3 = 3n \][/tex]
So, the [tex]\( n^{\text{th}} \)[/tex] term [tex]\( a_n \)[/tex] is [tex]\( 3n \)[/tex].
### Part (b): Sequence [tex]\(25, 22, 19, 16, \ldots\)[/tex]
This sequence is an arithmetic sequence where the first term [tex]\( a \)[/tex] is 25 and the common difference [tex]\( d \)[/tex] is -3.
The general formula for the [tex]\( n^{\text{th}} \)[/tex] term of an arithmetic sequence is:
[tex]\[ b_n = a + (n-1) \cdot d \][/tex]
For this sequence:
[tex]\[ b_n = 25 + (n-1) \cdot (-3) \][/tex]
Therefore, the [tex]\( n^{\text{th}} \)[/tex] term is:
[tex]\[ b_n = 25 - 3(n-1) = 25 - 3n + 3 = 28 - 3n \][/tex]
So, the [tex]\( n^{\text{th}} \)[/tex] term [tex]\( b_n \)[/tex] is [tex]\( 28 - 3n \)[/tex].
### Part (c): Sequence [tex]\(1, 4, 9, 16, \ldots\)[/tex]
This sequence is a quadratic sequence where each term is the square of its position in the sequence.
The general formula for the [tex]\( n^{\text{th}} \)[/tex] term of this quadratic sequence is:
[tex]\[ c_n = n^2 \][/tex]
So, the [tex]\( n^{\text{th}} \)[/tex] term [tex]\( c_n \)[/tex] is [tex]\( n^2 \)[/tex].
### Part (d): Sequence [tex]\(\frac{1}{3}, \frac{4}{5}, \frac{7}{7}, \frac{10}{9}, \ldots\)[/tex]
This sequence follows a pattern in both the numerator and the denominator.
Observe that the numerator increases by 3 for each subsequent term and the denominator increases by 2 for each subsequent term.
The general formula for the [tex]\( n^{\text{th}} \)[/tex] term of this sequence is:
[tex]\[ d_n = \frac{3n - 2}{2n + 1} \][/tex]
So, the [tex]\( n^{\text{th}} \)[/tex] term [tex]\( d_n \)[/tex] is [tex]\( \frac{3n - 2}{2n + 1} \)[/tex].
### Summary
- Part (a): [tex]\[ a_n = 3n \][/tex]
- Part (b): [tex]\[ b_n = 28 - 3n \][/tex]
- Part (c): [tex]\[ c_n = n^2 \][/tex]
- Part (d): [tex]\[ d_n = \frac{3n - 2}{2n + 1} \][/tex]
