Comprehensive worksheet designed to help students master the skill of finding the nth term in arithmetic sequences.

Problem: Finding Arithmetic nth Terms

The task involves finding the general formula for the \( n \)-th term of each arithmetic sequence. The general formula for the \( n \)-th term of an arithmetic sequence is:

\[
a_n = a + (n-1)d
\]

where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

#### Section A: Use the arithmetic sequences to find the \( n \)-th terms.

1. Sequence: 4, 7, 10, 13
- First term (\( a \)): 4
- Common difference (\( d \)): \( 7 - 4 = 3 \)
- \( n \)-th term: \( a_n = 4 + (n-1) \cdot 3 = 4 + 3n - 3 = 3n + 1 \)

2. Sequence: 6, 11, 16, 21
- First term (\( a \)): 6
- Common difference (\( d \)): \( 11 - 6 = 5 \)
- \( n \)-th term: \( a_n = 6 + (n-1) \cdot 5 = 6 + 5n - 5 = 5n + 1 \)

3. Sequence: 9, 11, 13, 15
- First term (\( a \)): 9
- Common difference (\( d \)): \( 11 - 9 = 2 \)
- \( n \)-th term: \( a_n = 9 + (n-1) \cdot 2 = 9 + 2n - 2 = 2n + 7 \)

4. Sequence: 15, 19, 23, 27
- First term (\( a \)): 15
- Common difference (\( d \)): \( 19 - 15 = 4 \)
- \( n \)-th term: \( a_n = 15 + (n-1) \cdot 4 = 15 + 4n - 4 = 4n + 11 \)

5. Sequence: 32, 42, 52, 62
- First term (\( a \)): 32
- Common difference (\( d \)): \( 42 - 32 = 10 \)
- \( n \)-th term: \( a_n = 32 + (n-1) \cdot 10 = 32 + 10n - 10 = 10n + 22 \)

6. Sequence: 76, 82, 88, 94
- First term (\( a \)): 76
- Common difference (\( d \)): \( 82 - 76 = 6 \)
- \( n \)-th term: \( a_n = 76 + (n-1) \cdot 6 = 76 + 6n - 6 = 6n + 70 \)

7. Sequence: 1, 4, 7, 10
- First term (\( a \)): 1
- Common difference (\( d \)): \( 4 - 1 = 3 \)
- \( n \)-th term: \( a_n = 1 + (n-1) \cdot 3 = 1 + 3n - 3 = 3n - 2 \)

8. Sequence: 3, 8, 13, 18
- First term (\( a \)): 3
- Common difference (\( d \)): \( 8 - 3 = 5 \)
- \( n \)-th term: \( a_n = 3 + (n-1) \cdot 5 = 3 + 5n - 5 = 5n - 2 \)

9. Sequence: -2, 0, 2, 4
- First term (\( a \)): -2
- Common difference (\( d \)): \( 0 - (-2) = 2 \)
- \( n \)-th term: \( a_n = -2 + (n-1) \cdot 2 = -2 + 2n - 2 = 2n - 4 \)

10. Sequence: -5, -2, 1, 4
- First term (\( a \)): -5
- Common difference (\( d \)): \( -2 - (-5) = 3 \)
- \( n \)-th term: \( a_n = -5 + (n-1) \cdot 3 = -5 + 3n - 3 = 3n - 8 \)

11. Sequence: -6, -1, 4, 9
- First term (\( a \)): -6
- Common difference (\( d \)): \( -1 - (-6) = 5 \)
- \( n \)-th term: \( a_n = -6 + (n-1) \cdot 5 = -6 + 5n - 5 = 5n - 11 \)

12. Sequence: -10, -7, -4, -1
- First term (\( a \)): -10
- Common difference (\( d \)): \( -7 - (-10) = 3 \)
- \( n \)-th term: \( a_n = -10 + (n-1) \cdot 3 = -10 + 3n - 3 = 3n - 13 \)

#### Section B: Find the \( n \)-th terms of the sequences.

1. Sequence: 4, 3, 2, 1
- First term (\( a \)): 4
- Common difference (\( d \)): \( 3 - 4 = -1 \)
- \( n \)-th term: \( a_n = 4 + (n-1) \cdot (-1) = 4 - n + 1 = 5 - n \)

2. Sequence: 8, 6, 4, 2
- First term (\( a \)): 8
- Common difference (\( d \)): \( 6 - 8 = -2 \)
- \( n \)-th term: \( a_n = 8 + (n-1) \cdot (-2) = 8 - 2n + 2 = 10 - 2n \)

3. Sequence: 17, 12, 7, 2
- First term (\( a \)): 17
- Common difference (\( d \)): \( 12 - 17 = -5 \)
- \( n \)-th term: \( a_n = 17 + (n-1) \cdot (-5) = 17 - 5n + 5 = 22 - 5n \)

4. Sequence: 0, -1, -2, -3
- First term (\( a \)): 0
- Common difference (\( d \)): \( -1 - 0 = -1 \)
- \( n \)-th term: \( a_n = 0 + (n-1) \cdot (-1) = 0 - n + 1 = 1 - n \)

5. Sequence: 1, -1, -3, -5
- First term (\( a \)): 1
- Common difference (\( d \)): \( -1 - 1 = -2 \)
- \( n \)-th term: \( a_n = 1 + (n-1) \cdot (-2) = 1 - 2n + 2 = 3 - 2n \)

6. Sequence: 3, -2, -7, -12
- First term (\( a \)): 3
- Common difference (\( d \)): \( -2 - 3 = -5 \)
- \( n \)-th term: \( a_n = 3 + (n-1) \cdot (-5) = 3 - 5n + 5 = 8 - 5n \)

7. Sequence: -4, -7, -10, -13
- First term (\( a \)): -4
- Common difference (\( d \)): \( -7 - (-4) = -3 \)
- \( n \)-th term: \( a_n = -4 + (n-1) \cdot (-3) = -4 - 3n + 3 = -3n - 1 \)

8. Sequence: -11, -13, -15, -17
- First term (\( a \)): -11
- Common difference (\( d \)): \( -13 - (-11) = -2 \)
- \( n \)-th term: \( a_n = -11 + (n-1) \cdot (-2) = -11 - 2n + 2 = -2n - 9 \)

9. Sequence: -16, -20, -24, -28
- First term (\( a \)): -16
- Common difference (\( d \)): \( -20 - (-16) = -4 \)
- \( n \)-th term: \( a_n = -16 + (n-1) \cdot (-4) = -16 - 4n + 4 = -4n - 12 \)

10. Sequence: 1.5, 3.5, 5.5, 7.5
- First term (\( a \)): 1.5
- Common difference (\( d \)): \( 3.5 - 1.5 = 2 \)
- \( n \)-th term: \( a_n = 1.5 + (n-1) \cdot 2 = 1.5 + 2n - 2 = 2n - 0.5 \)

11. Sequence: 8.8, 8.6, 8.4, 8.2
- First term (\( a \)): 8.8
- Common difference (\( d \)): \( 8.6 - 8.8 = -0.2 \)
- \( n \)-th term: \( a_n = 8.8 + (n-1) \cdot (-0.2) = 8.8 - 0.2n + 0.2 = 9 - 0.2n \)

12. Sequence: \(\frac{3}{4}, \frac{1}{2}, \frac{1}{4}, 0\)
- First term (\( a \)): \(\frac{3}{4}\)
- Common difference (\( d \)): \( \frac{1}{2} - \frac{3}{4} = \frac{2}{4} - \frac{3}{4} = -\frac{1}{4} \)
- \( n \)-th term: \( a_n = \frac{3}{4} + (n-1) \cdot \left(-\frac{1}{4}\right) = \frac{3}{4} - \frac{n}{4} + \frac{1}{4} = \frac{4}{4} - \frac{n}{4} = 1 - \frac{n}{4} \)

#### Extension: Create a sequence whose \( n \)-th term is in the following form:

1. \( an + b \)
- Example: \( 3n + 2 \)
- Sequence: 5, 8, 11, 14, ...

2. \( an - b \)
- Example: \( 3n - 2 \)
- Sequence: 1, 4, 7, 10, ...

3. \( -an + b \)
- Example: \( -3n + 5 \)
- Sequence: 2, -1, -4, -7, ...

4. \( -an - b \)
- Example: \( -3n - 2 \)
- Sequence: -5, -8, -11, -14, ...

Final Answer:

\[
\boxed{
\begin{array}{ll}
\text{Section A:} & \begin{aligned}
& 3n + 1, \quad 5n + 1, \quad 2n + 7, \quad 4n + 11, \quad 10n + 22, \quad 6n + 70, \\
& 3n - 2, \quad 5n - 2, \quad 2n - 4, \quad 3n - 8, \quad 5n - 11, \quad 3n - 13
\end{aligned} \\
\text{Section B:} & \begin{aligned}
& 5 - n, \quad 10 - 2n, \quad 22 - 5n, \quad 1 - n, \quad 3 - 2n, \quad 8 - 5n, \\
& -3n - 1, \quad -2n - 9, \quad -4n - 12, \quad 2n - 0.5, \quad 9 - 0.2n, \quad 1 - \frac{n}{4}
\end{aligned} \\
\text{Extension:} & \begin{aligned}
& 3n + 2, \quad 3n - 2, \quad -3n + 5, \quad -3n - 2
\end{aligned}
\end{array}
}
\]