This scenario represents an arithmetic sequence, where the starting salary is the first term, $a_1 = 3200$, and the monthly raise is the common difference, $d = 135$.

Step 1: General Term of the Arithmetic Sequence

The general term of an arithmetic sequence is given by:

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

Here:

• $a_1 = 3200$
• $d = 135$

Substitute the values into the formula:

$a_n = 3200 + (n-1) \cdot 135$

Simplify:

$a_n = 3200 + 135n - 135$

$a_n = 135n + 3065$

Thus, the general term of the sequence is:

$a_n = 135n + 3065$

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Step 2: Monthly Salary at the End of the Training

Assume "end of training" is after 12 months, so we need to find $a_{12}$. Use the general term:

$a_{12} = 135(12) + 3065$

Calculate:

$a_{12} = 1620 + 3065 = 4685$

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Final Results:

1. General term of the sequence: $a_n = 135n + 3065$
2. Monthly salary at the end of the training: $\$4685$

Would you like further clarification on this, or need help with related problems? Here are 5 follow-up questions for further exploration:

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1. How would the sequence change if the monthly raise were increased to $200?
2. What is the total amount Jose earns during his first year?
3. How many months would it take for Jose’s monthly salary to exceed $5000?
4. If the raises stop after the first year, what will Jose’s salary be after two years?
5. How does this arithmetic sequence compare to a geometric sequence in terms of salary growth?

Tip: Always verify the formula by testing it with the first few terms of the sequence!