Chapter 7.3, Problem 1MRPS

Solution

a.

To calculate : The rule that gives salary $a_n$ for the $n^{th}$ year of employment if a person accepts a job as an environmental engineer that pays a salary of $\$45,000$ in the first year. After that the salary increases by $3.5\%$ per year.

The rule that gives salary $a_n$ for the $n^{th}$ year of employment is $a_n=45000{\displaystyle \sum _{i=1}^{\infty }{\left(1.035\right)}^{i-1}}$ .

Given information :

A person accepts a job as an environmental engineer that pays a salary of $\$45,000$ in the first year. After that the salary increases by $3.5\%$ per year.

Formula used :

The $n^{th}$ term of a geometric series with first term as $a_1$ and common ratio $r$ is given by $a_n=a_1{r}^{n-1}$ .

Calculation :

Consider the salary in first year as $a_1=45,000$ .

Calculate the salary after the first year if the increase is $3.5\%$ .

  $\begin{array}{c}{a}_2=45,000\left(1+3.5\%\right)\\ =45000\left(1+\frac{3.5}{100}\right)\\ =45000\left(1.035\right)\end{array}$

Calculate the salary after the second year if the increase is $3.5\%$ .

  $\begin{array}{c}{a}_3=45,000\left(1+3.5\%\right)\left(1+3.5\%\right)\\ =45000\left(1+\frac{3.5}{100}\right)\left(1+\frac{3.5}{100}\right)\\ =45000{\left(1.035\right)}^2\end{array}$

The geometric series is written as $a_n=45000{\displaystyle \sum _{i=1}^{\infty }{\left(1.035\right)}^{i-1}}$ .

Thus, the rule that gives salary $a_n$ for the $n^{th}$ year of employment is $a_n=45000{\displaystyle \sum _{i=1}^{\infty }{\left(1.035\right)}^{i-1}}$ .

b.

To calculate : The salary during the $5^{th}$ year of employment if a person accepts a job as an environmental engineer that pays a salary of $\$45,000$ in the first year. After that the salary increases by $3.5\%$ per year.

The salary during the $5^{th}$ year of employment is $\$54638.53$ .

Given information :

A person accepts a job as an environmental engineer that pays a salary of $\$45,000$ in the first year. After that the salary increases by $3.5\%$ per year.

Formula used :

The $n^{th}$ term of a geometric series with first term as $a_1$ and common ratio $r$ is given by $a_n=a_1{r}^{n-1}$ .

Calculation :

Consider the rule that gives salary $a_n$ for the $n^{th}$ year of employment is $a_n=45000{\displaystyle \sum _{i=1}^{\infty }{\left(1.035\right)}^{i-1}}$ .

Calculate the fifth-year salary by substituting the value of $n$ as $5$ into the formula $a_n=45000{\displaystyle \sum _{i=1}^{\infty }{\left(1.035\right)}^{i-1}}$ .

  $\begin{array}{c}{a}_5=45000{\left(1.035\right)}^{5-1}\\ =45000{\left(1.035\right)}^4\\ =54638.53\end{array}$

Thus, the salary during the $5^{th}$ year of employment is $\$54638.53$ .

c.

To calculate : The total amount earned in 30 years if a person accepts a job as an environmental engineer that pays a salary of $\$45,000$ in the first year. After that the salary increases by $3.5\%$ per year.

The total salary for the $30$ year of employment is $\$2323020.48$ .

Given information :

A person accepts a job as an environmental engineer that pays a salary of $\$45,000$ in the first year. After that the salary increases by $3.5\%$ per year.

Formula used :

The sum of the first $n$ terms of geometric series with first term $a_1$ and common ration $r$ is calculated using the formula $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$ with $r\ne 1$ .

Calculation :

Calculate the sum of first $30$ years’ salary by substituting the value of $a_1$ as $\$45000$ , $n$ as $30$ and $r$ as $1.035$ into the formula $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$ .

  $\begin{array}{c}{S}_{30}=45000\left(\frac{1-{1.035}^{30}}{1-1.035}\right)\\ =45000\left(\frac{1-2.806}{-0.035}\right)\\ =45000\left(\frac{-1.806}{-0.035}\right)\\ =45000\left(54.62\right)\end{array}$

  $S_{30}=2323020.48$

Thus, the total salary for the $30$ year of employment is $\$2323020.48$ .