Problem 59 - Precalculus: Mathematical for Calculus

Question

Salary Increases A man gets a job with a salary of \(\$ 30,000\) a year. He is promised a \(\$ 2300\) raise each subsequent year. Find his total earnings for a 10 -year period.

Step-by-Step Solution

Answer

The total earnings over 10 years are \( \$403,500 \).

Step 1: Determine Annual Salary for Each Year

In the first year, his salary is \( \\(30,000 \). Each subsequent year, he receives a \( \\)2300 \) raise. Thus, his salary for the \( n^{th} \) year can be expressed as follows: \( 30000 + 2300 \times (n-1) \).

Step 2: List the Salaries Over 10 Years

Using the formula from Step 1, list the salary for each year for 10 years:\- Year 1: \( \\(30,000 \)- Year 2: \( 30,000 + 2300 \times 1 = \\)32,300 \)- Year 3: \( 30,000 + 2300 \times 2 = \\(34,600 \)- Year 4: \( 30,000 + 2300 \times 3 = \\)36,900 \)- Year 5: \( 30,000 + 2300 \times 4 = \\(39,200 \)- Year 6: \( 30,000 + 2300 \times 5 = \\)41,500 \)- Year 7: \( 30,000 + 2300 \times 6 = \\(43,800 \)- Year 8: \( 30,000 + 2300 \times 7 = \\)46,100 \)- Year 9: \( 30,000 + 2300 \times 8 = \\(48,400 \)- Year 10: \( 30,000 + 2300 \times 9 = \\)50,700 \).

Step 3: Apply the Sum Formula for an Arithmetic Sequence

The sequence of his salaries over 10 years is arithmetic with a first term \( a = 30,000 \), common difference \( d = 2300 \), and 10 terms. The sum \( S_n \) of an arithmetic sequence is given by \( S_n = \frac{n}{2} (2a + (n-1)d) \).

Step 4: Substitute Values into Sum Formula

Substitute \( n = 10 \), \( a = 30,000 \), and \( d = 2300 \) into the sum formula: \\[S_{10} = \frac{10}{2} (2 \times 30,000 + 9 \times 2300) \= 5 (60,000 + 20,700) \= 5 \times 80,700 \= 403,500 \\]

Step 5: Calculate Final Result

The total earnings over the 10-year period is \( \$403,500 \). This is obtained from the sum of the arithmetic sequence calculated in the previous step.

Key Concepts

Salary Calculation

Understanding how salaries increase over time is essential, especially when considering long-term financial planning. For this scenario, the initial salary is set at \( \\(30,000 \) per year. Importantly, each subsequent year offers a fixed raise of \( \\)2300 \). This setup creates a predictable pattern in the employee's earnings.

To find the yearly salary, you begin with the base salary plus the increases accrued over the years:

• Year 1: \( \\(30,000 \) (no raise yet as it’s the starting amount).
• Year 2: \( \\)30,000 + \\(2300 = \\)32,300 \).
• Year 3: \( \\(30,000 + 2 \times \\)2300 = \\(34,600 \).

This pattern is straightforward as we add \( \\)2300 \) for every additional year to the initial salary, tailoring the formula as: \( 30000 + 2300 \times (n-1) \) for the nth year.

This calculation is crucial for assessing how much one would earn annually, and it leads into understanding total earnings over a period, which is vital for savings and investment planning.

Sum Formula

The sum formula for an arithmetic sequence helps calculate the total earnings over multiple periods or years. Here, we're interested in the total earnings over 10 years.

The sequence forms an arithmetic progression with the initial salary as the first term and a steady increase – our common difference – being \( \\(2300\). To efficiently compute the total salary earned over 10 years without manually adding each year's salary, we apply the arithmetic sequence sum formula:

\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]

• \(n\): number of terms, which is 10 in this case.
• \(a\): first term, \(\\)30,000\).
• \(d\): common difference, \(\\(2300\).

It's a quick way to add up a series of salaries spanning many years. Inserting our values, we computed:

\[ S_{10} = \frac{10}{2} \times (2 \times 30,000 + 9 \times 2300) = 5 \times 80,700 = 403,500 \]

The sum comes out to \(\\)403,500\), revealing the total earnings from the job over 10 years.

Common Difference

The notion of a common difference is a cornerstone in understanding arithmetic sequences, particularly in this scenario where salary raises are consistent each year.

In arithmetic sequences, the common difference \( d \) is a constant. It represents the uniform increase from one term to the next. Here, the common difference is \( \\(2300 \), marking the amount added to the salary annually.

• The first term should be the initial salary, \( \\)30,000 \).
• Each year after that, this common difference increases the salary by \( \$2300 \).

Understanding and identifying the common difference allows for constructing future salaries over time. This becomes crucial when forecasting for personal finance, knowing how much increase to expect, and planning accordingly.

Additionally, by recognizing the common difference, constructing or calculating the total sum over different periods becomes a logical extension, making comprehensive salary forecasting more intuitively comprehensible.