Chapter 12.2, Problem 43E

a.

To determine

To calculate:

Payments will be at the end of the ${10}^{th},{20}^{th},{40}^{th}$

Answer to Problem 43E

Payments will be at the end of the ${10}^{th},{20}^{th},{40}^{th}$ are $\$11.79,\$30.58,\$205.72$ .

Explanation of Solution

Given information:

First term $a_1$ in the series is the payment of the buyer in first week.

  $a_1=\$5,a_2=\$5.50$

Calculation:

The common ratio of this geometric series is,

  $\begin{array}{l}r=\frac{a_2}{a_1}\\ r=\frac{5.50}{5}\\ r=1.1\end{array}$

Apply the formula for the $n^{th}$ term of a geometric series.

  $a_n=a_1{r}^{n-1}$

Payment after ${10}^{th}$ week is,

  $\begin{array}{l}{a}_{10}=a_1{r}^{10-1}\\ a_{10}=\left(\$5\right){\left(1.1\right)}^{10-1}\\ a_{10}=\$11.79\end{array}$

Payment after $20$ week is,

  $\begin{array}{l}{a}_{20}=a_1{r}^{20-1}\\ a_{20}=\left(\$5\right){\left(1.1\right)}^{20-1}\\ a_{20}=\$30.58\end{array}$

Payment after $20$ week is,

  $\begin{array}{l}{a}_{40}=a_1{r}^{40-1}\\ a_{40}=\left(\$5\right){\left(1.1\right)}^{40-1}\\ a_{40}=\$205.72\end{array}$

Thus, Payments will be at the end of the ${10}^{th},{20}^{th},{40}^{th}$ are $\$11.79,\$30.58,\$205.72$ .

b.

To determine

To calculate:

Total cost of TV.

Answer to Problem 43E

Total cost of TV is $\$7052.15$ .

Explanation of Solution

Given information:

First term $a_1$ in the series is the payment of the buyer in first week.

  $a_1=\$5,a_2=\$5.50$

Calculation:

Apply the formula for the sum of a geometric series,

  $S_n=\frac{a_1-a_1{r}^n}{1-r}$

For one year there are $52$ weeks are

  $\begin{array}{l}{S}_n=\frac{\left(\$5\right)-\left(\$5\right){\left(1.1\right)}^{52}}{1-1.1}\\ S_n=\$7052.15\end{array}$

Thus, Total cost of TV is $\$7052.15$ .

c.

To determine

To explain:

Reason for Total cost of TV found in pat (b) is not accurate..

Answer to Problem 43E

Sum of the payments will actually be more than the sum found in part (b).

Explanation of Solution

Each payment made is rounded to the nearest penny so the sum of the payments will actually be more than the sum found in part (b).