Chapter 8, Problem 19Q

a.

To determine

Find the rule for the salary $a_n$ ( $n\text{th}$ year of employment).

Answer to Problem 19Q

The rule for the salary $a_n$ ( $n\text{th}$ year of employment) is $a_n=a_1\cdot {1.035}^{n-1}$ .

Explanation of Solution

Given:

The salary for the first year is $\$45,000$ and the salary increases by $3.5\%$ per year.

Calculation:

First year salary $a_1=45,000$

Second year salary $a_2=45000\cdot 1.035$

Third year salary $a_3=45000\cdot 1.035\cdot 1.035=45000\cdot {1.035}^2$

  $\begin{array}{l}\cdot \cdot \cdot \\ \cdot \cdot \cdot \\ \cdot \cdot \cdot \end{array}$

This pattern shows the geometric sequence.

The $n\text{th}$ year salary $a_n=45000\cdot {1.035}^{n-1}=a_1\cdot {1.035}^{n-1}$

Hence the rule for the salary $a_n$ ( $n\text{th}$ year of employment)is $a_n=a_1\cdot {1.035}^{n-1}$ .

b.

To determine

Find the salary for $5\text{th}$ year of employment.

Answer to Problem 19Q

The fifth year salary is $\$51,638.535$ .

Explanation of Solution

Given:

The salary for the first year is $\$45,000$ and the salary increases by $3.5\%$ per year.

Calculation:

First year salary $a_1=45,000$

Second year salary $a_2=45000\cdot 1.035$

Third year salary $a_3=45000\cdot 1.035\cdot 1.035=45000\cdot {1.035}^2$

  $\begin{array}{l}\cdot \cdot \cdot \\ \cdot \cdot \cdot \\ \cdot \cdot \cdot \end{array}$

This pattern shows the geometric sequence.

The $n\text{th}$ year salary $a_n=45000\cdot {1.035}^{n-1}=a_1\cdot {1.035}^{n-1}$

Fifth year salary

  $\begin{array}{l}{a}_5=a_1\cdot {1.035}^{5-1}\\ =45000\cdot {1.035}^4\\ =51638.535\end{array}$

Hence the fifth year salary is $\$51,638.535$ .

c.

To determine

Find the totalearningsin ten years.

Answer to Problem 19Q

The total earnings in ten years are $\$527912.69$ .

Explanation of Solution

Given:

The salary for the first year is $\$45,000$ and the salary increases by $3.5\%$ per year.

Calculation:

First year salary $a_1=45,000$

Second year salary $a_2=45000\cdot 1.035$

Third year salary $a_3=45000\cdot 1.035\cdot 1.035=45000\cdot {1.035}^2$

  $\begin{array}{l}\cdot \cdot \cdot \\ \cdot \cdot \cdot \\ \cdot \cdot \cdot \end{array}$

This pattern shows the geometric sequence.

The $n\text{th}$ year salary $a_n=45000\cdot {1.035}^{n-1}=a_1\cdot {1.035}^{n-1}$

Use geometric sum formula for the total salary.

  $S_n=\frac{a_1\left(1-r^n\right)}{1-r},r\ne 1$

  $a_1=45000,r=1.035,n=10$

  $\begin{array}{l}{S}_{10}=\frac{45000\left[1-{\left(1.035\right)}^{10}\right]}{1-1.035}\\ =\frac{45000\cdot \left(-0.41\right)}{-0.035}\\ =\frac{18476.9}{0.035}\\ =527912.69\end{array}$

Hence the totalearnings in ten yearsare $\$527912.69$ .