Chapter 12.2, Problem 63E

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.

To determine

To find: The total earning for a 10-year period.

Answer to Problem 63E

The total earning for 10-years is $403,500.

Explanation of Solution

The salary of a man rises by some fixed amount every year.

Here the arithmetic regression is increasing.

Given:

Salary in first year is $30,000.

Increase in the salary by $2300.

Total years are 10.

Formulas used:

The nth partial sum of an arithmetic sequence is,

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

Calculation:

For the first year salary is $30,000, without any raise.

So, first term of an arithmetic progression is 30000.

From the second year, the salary increases by $2300.

So, the common difference is 2300.

And total years are 10, so, value of n is 10.

Substitute 30000 for a, 2300 for d and 10 for n in the formula above to find the total earning,

$\begin{array}{c}{S}_n=\frac{10}{2}\left[2\times 30000+\left(10-1\right)2300\right]\\ =5\left(60000+20700\right)\\ =5\times 80700\\ =403500\end{array}$

Thus, the total earning for 10-years is $403,500.