Chapter 4.6, Problem 54E

(a)

To determine

To explain: the graph represents an arithmetic sequence.

Answer to Problem 54E

The sequence is arithmetic.

Explanation of Solution

Given:

  

Calculation:

Now, from the graph one can write the terms as the function of cost in dollars to the size of advertisement in pages

  $f\left(n\right)=a_1+\left(n-1\right)d$ Where d is difference and $a_1$ is the first cost

So, from the graph the terms are

  $f\left(1\right)=20000,f\left(2\right)=32000,f\left(3\right)=45000\text{ and }f\left(4\right)\text{=}58000$

Now consider the above terms to check the sequence is arithmetic or not

So, check the difference between the terms

  $\begin{array}{l}d=f\left(2\right)-f\left(1\right)\hfill \\ =32500-20000\hfill \\ =12500\hfill \end{array}$

Similarly,

  $\begin{array}{l}d=f\left(3\right)-f\left(2\right)\hfill \\ =45000-32500\hfill \\ =12500\hfill \end{array}$

And

  $\begin{array}{l}d=f\left(4\right)-f\left(3\right)\hfill \\ =57500-45000\hfill \\ =12500\hfill \end{array}$

The difference between the terms is same.

Conclusion:

Therefore, the sequence is arithmetic.

(b)

To determine

To explain: the estimation of the cost of a six-page advertisement in the magazine.

Answer to Problem 54E

The cost of the six-page advertisement can be estimated using the function representing the cost over the pages, and the cost is 82500 dollars.

Explanation of Solution

Calculation:

To estimate the cost of the six-page advertisement in the magazine, write the function

Consider the function $f\left(n\right)=a_1+\left(n-1\right)d$ and the terms $a_1=20000$ , $d=12500$ and $n=6$

  $\begin{array}{l}f\left(6\right)=20000+\left(6-1\right)\left(12500\right)\hfill \\ \begin{array}{l}=20000+\left(5\right)\left(12500\right)\\ =82500\end{array}\hfill \end{array}$

Conclusion:

Therefore, the cost of the six-page advertisement can estimate using the function representing the cost over the pages, and the cost is 82500 dollars.