Chapter 12, Problem 5P

(a)

To determine

To explain: The sequence that models the amount $A_n$ of the pollutant in the lake at the end of the nth year is $A_n=0.30A_{n-1}+2400$.

Explanation of Solution

Formula to calculate the amount of pollutants at the end of the nth year is,

$\left[\begin{array}{l}\text{Amount}\text{of}\text{the}\\ \text{pollutants}\text{at}\\ \text{end}\text{of}n\text{th}\text{year}\end{array}\right]=\left[\begin{array}{l}\text{Amount}\text{of}\text{the}\\ \text{pollutants}\text{at}\text{the}\\ \text{end}\text{of}\text{last}\text{year}\end{array}\right]+\left[\begin{array}{l}\text{rate}\text{of}\text{pollutants}\\ \text{per}\text{year}\end{array}\right]\times \left[\begin{array}{l}\text{Amount}\text{of}\text{the}\\ \text{pollutants}\text{at}\text{the}\\ \text{end}\text{of}\text{last}\text{day}\end{array}\right]$

Since, the rate of pollutants which are expelled is 70%. Thus the rate of pollutants which are not expelled is $30\%$.

Assume $A_0$ be the amount of initial pollutants and $A_1$ be the amount of pollutants at the end of first year and so on.

The amount of pollutants at the end of the nth year is $A_n$ and the rate of pollutants is 30% per year.

Amount of chemical present in the lake at the beginning time is 2400 tons.

The recursive sequence that models the amount $A_n$ at the end of the nth year is given by,

$A_n=0.03A_{n-1}+2400$ (1)

Hence, the sequence that models the amount $A_n$ of the pollutant in the lake at the end of the nth year is $A_n=0.30A_{n-1}+2400$.

(b)

To determine

To find: The first five terms of the sequence $A_n$.

Answer to Problem 5P

The first five terms of the sequence $A_n$ are $2400$, $2472$, $2474.16$, $2474.22$ and $2474.23$.

Explanation of Solution

Calculation:

Since the initial amount of pollutants present in the lake at the beginning is 2400 tons. Thus, $A_0=2400$.

Substitute 1 for $n$ in equation (1) from part (a), to evaluate $A_1$.

$\begin{array}{c}{A}_1=0.03A_0+2400\\ =0.03\left(2400\right)+2400\left[\because A_0=2400\right]\\ =2400\left(1+0.03\right)\\ =2472\end{array}$

Substitute 2 for $n$ in equation (1) from part (a), to evaluate $A_2$.

$\begin{array}{c}{A}_2=0.03A_1+2400\\ =0.03\left(0.03\left(2400\right)+2400\right)+2400\left[\because A_1=0.03\left(2400\right)+2400\right]\\ =2400\left({\left(0.03\right)}^2+0.03+1\right)\\ =2474.16\end{array}$

Substitute 3 for $n$ in equation (1) from part (a), to evaluate $A_3$.

$\begin{array}{c}{A}_3=0.03A_2+2400\\ =0.03\left(2400\left({\left(0.03\right)}^2+0.03+1\right)\right)+2400\left[\because A_2=2400\left({\left(0.03\right)}^2+0.03+1\right)\right]\\ =2400\left({0.03}^3+{0.03}^2+0.03+1\right)\\ =2474.22\end{array}$

Substitute 4 for $n$ in equation (1) from part (a), to evaluate $A_4$.

$\begin{array}{c}{A}_4=0.03A_3+2400\\ =0.03\left(2400\left({0.03}^3+{0.03}^2+0.03+1\right)\right)+2400\\ =2400\left({0.03}^4+{0.03}^3+{0.03}^2+0.03+1\right)\\ =2474.23\end{array}$

Hence, the first five terms of the sequence $A_n$ are $2400$, $2472$, $2474.16$, $2474.22$ and $2474.23$.

(c)

To determine

To find: The formula for $A_n$ in terms of $n$.

Answer to Problem 5P

The value of $A_n$ is. $1028.57\left(\left(1-{0.03}^n\right)\right)+2400$.

Explanation of Solution

Calculation:

Since, the terms $A_1,A_2,A_3,\mathrm{...}$ in the part (b) follows the same pattern.

The general term $A_n$ can be written as,

$\begin{array}{c}{A}_n=2400\left({0.03}^n+{0.03}^{n-1}+\cdots +{0.03}^2+0.03+1\right)\\ =2400\left({0.03}^n+{0.03}^{n-1}+\cdots +{0.03}^2+0.03\right)+2400\\ =2400\left({\displaystyle \sum _{i=1}^n{0.03}^i}\right)+2400\end{array}$ (2)

Since ${0.03}^i,i=1,2,\mathrm{....},n$ is the geometric sequence with first term $0.03$ and the common ratio $0.03$.

The sum of the geometric sequence is,

$S_n=a\left(\frac{1-r^n}{1-r}\right)$

Substitute $0.03$ for a and $0.03$ for r in the above formula.

$\begin{array}{c}{S}_n=0.03\left(\frac{1-{0.03}^n}{1-0.03}\right)\\ =0.03\left(\frac{1-{0.03}^n}{0.07}\right)\\ =\frac{3}{7}\left(1-{0.03}^n\right)\end{array}$

Substitute $\frac{3}{7}\left(1-{0.03}^n\right)$ for ${\displaystyle \sum _{i=1}^n{0.03}^i}$ in equation (2) to evaluate $A_n$ as follows,

$\begin{array}{c}{A}_n=2400\left(\frac{3}{7}\left(1-{0.03}^n\right)\right)+2400\\ =1028.57\left(\left(1-{0.03}^n\right)\right)+2400\end{array}$ (3)

Hence, the value of $A_n$ is $1028.57\left(\left(1-{0.03}^n\right)\right)+2400$.

(d)

To determine

To find: The pollutants remain in the lake after 6 year and after a long time.

Answer to Problem 5P

The pollutants remain in the lake after 6 year is $3427.82$ and after a long time is $3427.82$.

Explanation of Solution

Calculation:

For the pollutants remains in the lake after 6 year, substitute 6 for n in equation (3) to evaluate $A_6$.

$\begin{array}{c}{A}_6=1028.57\left(\left(1-{0.03}^6\right)\right)+2400\\ =3427.82\end{array}$

For the pollutants remains in the lake after a long time, substitute $\infty$ for n in equation (2) to evaluate $A_{\infty }$.

$\begin{array}{c}{A}_{\infty }=2400\left({\displaystyle \sum _{i=1}^{}{0.03}^i}\right)+2400\\ =2400\left(\frac{0.03}{1-0.03}\right)+2400\\ =2400\left(\frac{3}{7}\right)+2400\\ =3427.82\end{array}$

Hence, the pollutants remain in the lake after 6 year is $3427.82$ and after a long time is $3427.82$.

(e)

To determine

To sketch: The graph of the given equation $A_n=0.30A_{n-1}+2400$.

Explanation of Solution

The value of $A_0=2400$.

Substitute some values of n and find the corresponding values of the function as given below in the table,

$n$	$A_n$
0	2400
1	2472
2	2474.16
3	2474.22
4	2474.23
5	3427.82
6	3427.82
7	3427.82
8	3727.82

Table

Plot the points in the graph and connect the points.

From the above figure, it can be observed that as the value of n increases in $\left[6,\infty \right)$ the corresponding value of $A_n$ remains constant.