Chapter 8.2, Problem 77E

To determine

To find: the total profit during the first 5 years of business.

Answer to Problem 77E

Total profit during the first 5 years is $200000.

Explanation of Solution

Given information:

A hardware store makes a profit of $30000 during its first year. The store owner sets a goal of increasing profits by $5000 given year for 4 years.

Concept used:

An arithmetic sequence of n terms, has the form $a_1,a_2,a_3,\mathrm{...},a_n$ and should have a common difference between each two consecutive terms.

That is

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}$

Common difference can be defined by d .

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}=d$

nth term of the arithmetic sequence has the form

  $a_n=a_1+\left(n-1\right)d$

Where $a_1$ is the first term of the sequence, and d is the common difference.

Sum of an arithmetic finite sequence has the form

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

Here, n is number of terms, $a_1$ is the first term of the sequence, and $a_n$ is the last term of the sequence.

Calculation:

A hardware makes a profit per year is $30000 and increasing profit $5000 per year.

The profit of list year $a_1=\$30000$

Profit of $n$ year

  $\begin{array}{c}{a}_n=30000+\left(n-1\right)5000\\ =5000n+25000\end{array}$

Profit of fifth year by using formula

  $\begin{array}{c}{a}_n=a+\left(n-1\right)d\\ a_5=30000+\left(5-1\right)5000\\ =30000+4\times 5000\\ =30000+20000\\ =50000\end{array}$

Total profit during the first 5 years by using formula

  $\begin{array}{c}{S}_n=\frac{n}{2}\left(a_1+a_n\right)\\ S_5=\frac{5}{2}\left(30000+50000\right)\\ =\frac{5}{2}\times 80000\\ =5\times 40000\\ =200000\end{array}$

Hence, total profit during the first 5 years is $200000.