We can approach this problem as an arithmetic sequence where the first term $a_1$ represents Jose's starting salary, and the common difference $d$ is the amount of the monthly raise. Let's work through the solution.

Part 1: General Term of the Arithmetic Sequence

The formula for the general term of an arithmetic sequence is:

$a_n = a_1 + (n - 1) \cdot d$

where:

• $a_1 = 2500$ (starting salary),
• $d = 100$ (monthly raise),
• $n$ represents the month number.

Thus, the general term $a_n$ of this sequence is:

$a_n = 2500 + (n - 1) \cdot 100$

Part 2: Jose's Monthly Salary at the End of His Training

Assuming his training lasts for 12 months (1 year), we substitute $n = 12$ into the formula for $a_n$:

$a_{12} = 2500 + (12 - 1) \cdot 100$ $a_{12} = 2500 + 11 \cdot 100$ $a_{12} = 2500 + 1100 = 3600$

So, Jose's monthly salary at the end of his training will be $3600.

Summary:

• General term of the sequence: $a_n = 2500 + (n - 1) \cdot 100$
• Jose's monthly salary at the end of his training: $3600

Do you need further details on this? Here are 5 related questions for practice:

1. What would Jose’s salary be in the 6th month of his training?
2. How much total salary would Jose have earned by the end of 12 months?
3. If the raise was instead $150 per month, what would be his salary at the end of 12 months?
4. What is the sum of the first 12 terms of this arithmetic sequence?
5. How would the formula change if the raise began in the second month instead of the first?

Tip: For arithmetic sequences, you can quickly find the $n$-th term using the general term formula—especially useful for salary-based questions!