Chapter 7.3, Problem 61PS

Solution

i.

To identify: The rules giving the salaries a and b, in the nth year at companies A and B. respectively. Tell whether the sequence represented by each rule is arithmetic, geometric, or neither.

Two companies, company A and company B, offer the same starting salary of $20,000 per year. Company A gives a raise of $1000 each year. Company B gives a raise of 4% each year.

Company A is arithmetic and company B is geometric.

Given information:

The given terms are.

  $\begin{array}{l}{a}_1=20000\text{and}d=10000\\ b_1=20000\text{and}r=1+0.04\end{array}$

Explanation:

  $a_n$ = the sequence representing the salary at company A

  $b_n$ = denotes the sequence that models the salary at company B.

Put the value of $a_1=20000andd=10000$ into $a_n=a+\left(n-1\right)d$

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ a_n=20000+\left(n-1\right)10000\end{array}$

Because the salary at company A is raised by the same amount each year, the sequence is arithmetic.

Put the value of $b_1=20000\text{and}r=1.04$ into $b_n=b_1{r}^{n-1}$

  $\begin{array}{l}{b}_n=b_1{r}^{n-1}\\ b_n=20000{\left(1.04\right)}^{n-1}\end{array}$

Because the salary at company B is raised by the same percentage each year, the sequence is geometric.

Therefore company A is arithmetic and company B is geometric.

ii.

To identify: Drawing Graphs Graph each sequence in the same coordinate plane.

The value of the sequence and plot the point on graph.

Given information:

The given terms are.

  $\begin{array}{l}{a}_1=20000\text{and}d=10000\\ b_1=20000\text{and}r=1+0.04\end{array}$

Explanation:

Consider the given sequence.

  $\begin{array}{l}{a}_n=20000+\left(n-1\right)10000\\ b_n=20000{\left(1.04\right)}^{n-1}\end{array}$

From above explanation.

Draw the table calculate the value of $a_n\text{and }{b}_n$

  $\begin{array}{ccc}n& a_n& b_n\\ 1& 20000& 20000\\ 2& 21000& 20800\\ 3& 22000& 21632\\ 4& 23000& 22497.28\\ 5& 24000& 23397.171\\ 6& 25000& 24333.058\\ 7& 26000& 25306.38\\ 8& 27000& 26318.636\\ 9& 28000& 27371.381\end{array}$

Draw the graph following the table value.

  

iii.

To calculate: Sums For each company, find the sum of wages earned during the first 20 years of employment

Company A $\$590000$

Company B $\approx \$595562$

Given information:

The given terms are.

  $\begin{array}{l}{a}_1=20000\text{and}d=10000\\ b_1=20000\text{and}r=1.04\end{array}$

Calculaation:

  $a_n$ = the sequence representing the salary at company A

  $b_n$ = denotes the sequence that models the salary at company B.

Put the value of $a_1=20000,n=20\text{and}d=10000$ into $a_n=a+\left(n-1\right)d$

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ a_n=20000+\left(20-1\right)10000\\ a_n=39000\end{array}$

Put the value of $a_1=20000,n=20\text{and}{a}_n=39000$ into $s_n=\frac{n\left(a_1+a_n\right)}{2}$

  $\begin{array}{l}{s}_n=\frac{n\left(a_1+a_n\right)}{2}\\ s_n=\frac{20\left(20000+39000\right)}{2}\\ s_n=590000\end{array}$

Put the value of $b_1=20000,n=20\text{and}r=1.04$ into $T_n=\frac{a_1\left(1-r^n\right)}{1-r}$

  $\begin{array}{l}{T}_n=\frac{a_1\left(1-r^n\right)}{1-r}\\ T_n=\frac{20000\left(1-{1.04}^{20}\right)}{1-1.04}\\ T_n\approx 595562\end{array}$

iv.

To identify: How many years the total amount earned at company B is greater than the total amount earned at company A. Use a graphing calculator or spreadsheet

The required years are $19$ .

Given information:

The given terms are.

  $\begin{array}{l}{a}_1=20000\text{and}d=10000\\ b_1=20000\text{and}r=1.04\end{array}$

Explanation:

  $a_n$ = the sequence representing the salary at company A

  $b_n$ = denotes the sequence that models the salary at company B.

See that the first value of $n$ such that sum of the sequence $b_n$ is greater than the sum of the sequence $a_n$ is $n=19$ .

It mean that the first year such that the total amount earned at company B is greater than the total amount earned at company A is the $19th$ year

  

So, $19$ year is the required answer.