Consider the following sequences:

3n-2: 1, 4, 7, 10, 13, 16, 19, 22, 25, ...

2n+1: 3, 5, 7, 9, 11, 13, 15, 17, 19, ...

The common terms of these two sequences are 7, 13, 19, which is a sequence generated by the rule 6n+1.

Questions:

How does that rule 6n+1 relate to the rules 3n-2 and 2n+1?
Given the rules of two other arithmetic sequences, can we derive the rule for the sequence that generates the terms that are common to both sequences?

### Relationship Between the Rules 3n-2, 2n+1, and 6n+1

Let's analyze how the sequence generated by the rule \( 6n + 1 \) is related to the sequences generated by \( 3n - 2 \) and \( 2n + 1 \).

#### Step 1: Set Up the Equations

Given the two sequences:
1. \( a_n = 3n - 2 \)
2. \( b_m = 2m + 1 \)

We want to find the values of \( n \) and \( m \) for which \( a_n = b_m \). That is, we need to solve the equation:
\[
3n - 2 = 2m + 1
\]

#### Step 2: Solve for \( n \) and \( m \)

Let's rearrange the equation:
\[
3n - 2 = 2m + 1
\]
\[
3n = 2m + 3
\]
\[
n = \frac{2m + 3}{3}
\]

For \( n \) to be an integer, \( 2m + 3 \) must be divisible by 3. Therefore:
\[
2m + 3 \equiv 0 \ (\text{mod} \ 3)
\]
This simplifies to:
\[
2m \equiv 0 \ (\text{mod} \ 3)
\]
Since 2 and 3 are coprime, this means:
\[
m \equiv 0 \ (\text{mod} \ 3)
\]
So \( m \) must be a multiple of 3, say \( m = 3k \) for some integer \( k \).

Substituting \( m = 3k \) back into the equation for \( n \):
\[
n = \frac{2(3k) + 3}{3} = \frac{6k + 3}{3} = 2k + 1
\]

So the common terms occur when \( n = 2k + 1 \) and \( m = 3k \).

#### Step 3: Find the Common Sequence Rule

Substitute \( k \) into either sequence to find the common sequence:
- For \( a_n = 3n - 2 \) with \( n = 2k + 1 \):
\[
a_{2k+1} = 3(2k + 1) - 2 = 6k + 3 - 2 = 6k + 1
\]
- For \( b_m = 2m + 1 \) with \( m = 3k \):
\[
b_{3k} = 2(3k) + 1 = 6k + 1
\]

So, the common sequence is generated by the rule \( 6k + 1 \), which can be written as \( 6n + 1 \) where \( n = k \).

### Generalization

Given two arithmetic sequences \( a_n = p_1n + q_1 \) and \( b_m = p_2m + q_2 \), we can derive the rule for the sequence of common terms by setting \( a_n = b_m \) and solving for \( n \) and \( m \). The resulting equation will help us determine the form of the common sequence. This process generally involves finding integer solutions for \( n \) and \( m \) that satisfy the equality condition for the two sequences.

### Summary

The rule \( 6n + 1 \) for the common sequence is derived from the intersection of the sequences defined by \( 3n - 2 \) and \( 2n + 1 \). The general method involves equating the two sequence rules and solving for integer values of \( n \) and \( m \), which leads to the discovery of the common sequence.

### Follow-Up Questions
1. What happens if the difference between the two sequence rules is not constant?
2. How does the general formula for the common sequence change if one sequence has a non-linear rule?
3. Can you apply the same method to sequences with different arithmetic patterns, such as \( 5n + 2 \) and \( 7n - 3 \)?
4. How does this method apply when the sequences have the same linear coefficient but different constants?
5. What are the implications of this method for finding common terms in sequences involving higher-order polynomials?

**Tip:** Always check the divisibility condition when finding common terms in sequences, as it can provide insights into the general form of the solution.

Explore the relationship between arithmetic sequences 3n-2, 2n+1, and the derived sequence 6n+1. Learn how to find common terms in sequences using mathematical formulas and equations. Suitable for high school students studying arithmetic sequences and linear equations.

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