Chapter 9.2, Problem 70E

a.

To determine

the salary during the sixth year of employment.

Answer to Problem 70E

  $\$45,550$

Explanation of Solution

Concept Used:

The n^th term of an arithmetic sequence with first term $a_1$ and common difference d is given by:

  $a_n=a_1+\left(n-1\right)d$

Given problem represents an arithmetic sequence with first term $a_1=\$36,800$ and $d=\$1750$ .

Now the n^th term of this sequence is given by,

  $\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ =36800+\left(n-1\right)\left(1750\right)\\ =36800+1750n-1750\\ =35050+1750n\end{array}$

The salary during the sixth year of employment is the sixth term of this sequence.

  $\begin{array}{c}{a}_6=35050+1750\left(6\right)\\ =35050+10500\\ =\$45,550\end{array}$

Conclusion:

The salary during the sixth year of employment is $\$45,550$ .

b.

To determine

the total compensation from the company through six full years of employment.

Answer to Problem 70E

  $\$247,050$

Explanation of Solution

Concept Used:

The sum of first n terms of an arithmetic sequence with first term $a_1$ and last term $a_n$ is given by:

  $S_n=\frac{n}{2}\left(a_1+a_n\right)$

Given problem represents an arithmetic sequence with first term $a_1=\$36,800$ and $d=\$1750$ .

The total compensation from the company through six full years of employment is the sum of first 6 terms of this sequence which is given by:

  $\begin{array}{c}{S}_6=\frac{6}{2}\left(a_1+a_6\right)\\ =3\left(36,800+45,550\right)\\ =3\left(82,350\right)\\ =\$247,050\end{array}$

Conclusion:

The total compensation from the company through six full years of employment is $\$247,050$ .