Chapter 12, Problem 1P

(a)

To determine

The recursive sequence that models the amount $A_n$ at the end of the nth day of 2015.

Answer to Problem 1P

The recursive sequence that models the amount $A_n$ at the end of the nth day of 2015 is $A_n=1.0001A_{n-1}$.

Explanation of Solution

Given:

Initial amount deposited at the start of 2015 is 275,000 and the rate of interest is $3.65\%\text{per}\text{year}$.

Formula used:

Formula to calculate the amount at the end of the nth day is,

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

Calculation:

The total number of days in a year is 365.

The rate of interest per day is calculated by the formula,

$\begin{array}{c}\text{Rate}\text{of}\text{interest}\text{per}\text{day}=\frac{\text{Rate}\text{of}\text{interest}\text{per}\text{year}}{365}\\ =\frac{3.65}{365}\left(\text{Rate}\text{of}\text{interest}=3.65\%\right)\\ =0.1\%\end{array}$

Rate of interest is $0.01\%\text{per}\text{day}$.

Assume $A_0$ be the original deposit and $A_1$ be the amount at the end of first day and so on. The amount at the end of the nth day of 2015 is $A_n$ and the rate of interest is $0.01\%$. Thus the recursive sequence that models the amount $A_n$ at the end of the nth day of 2015 is given as,

$\begin{array}{c}{A}_n=A_{n-1}+0.0001A_{n-1}\\ A_n=1.0001A_{n-1}\end{array}$ (1)

Hence, the recursive sequence that models the amount $A_n$ at the end of the nth day of 2015 is $A_n=1.0001A_{n-1}$.

(b)

To determine

The first eight terms of the sequence $A_n$ to the nearest cent.

Answer to Problem 1P

The first eight terms of the sequence $A_n$ to the nearest cent are $275000$, $275027.5$, $275055$, $275027.5$, $275110.02$, $275137.53$, $275165.04$, $275165.04$.

Explanation of Solution

Given:

Initial amount deposited at the start of 2015 is $275000$ and the rate of interest is $3.65\%\text{per}\text{year}$.

Calculation:

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

$\begin{array}{c}{A}_1=1.001A_0\\ =1.0001\left(275000\right)\left[\because A_0=275000\right]\\ =275027.5\end{array}$

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

$\begin{array}{c}{A}_2=1.001A_1\\ =1.0001\left(1.0001\left(275000\right)\right)\left[\because A_1=1.0001\left(275000\right)\right]\\ ={\left(1.0001\right)}^2\left(275000\right)\\ =275055\end{array}$

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

$\begin{array}{c}{A}_3=1.001A_2\\ =1.0001\left({\left(1.0001\right)}^2\left(275000\right)\right)\left[\because A_2={\left(1.0001\right)}^2\left(275000\right)\right]\\ ={\left(1.0001\right)}^3\left(275000\right)\\ =275027.5\end{array}$

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

$\begin{array}{c}{A}_4=1.001A_3\\ =1.0001\left({\left(1.0001\right)}^3\left(275000\right)\right)\left[\because A_3={\left(1.0001\right)}^3\left(275000\right)\right]\\ ={\left(1.0001\right)}^4\left(275000\right)\\ =275110.02\end{array}$

Substitute 5 for $n$ in equation (1) from part (a), to evaluate $A_5$.

$\begin{array}{c}{A}_5=1.001A_4\\ =1.0001\left({\left(1.0001\right)}^4\left(275000\right)\right)\left[\because A_4={\left(1.0001\right)}^4\left(275000\right)\right]\\ ={\left(1.0001\right)}^5\left(275000\right)\\ =275137.53\end{array}$

Substitute 6 for $n$ in equation (1) from part (a), to evaluate $A_6$.

$\begin{array}{c}{A}_6=1.001A_5\\ =1.0001\left({\left(1.0001\right)}^5\left(275000\right)\right)\left[\because A_5={\left(1.0001\right)}^5\left(275000\right)\right]\\ ={\left(1.0001\right)}^6\left(275000\right)\\ =275165.04\end{array}$

Substitute 7 for $n$ in equation (1) from part (a), to evaluate $A_7$.

$\begin{array}{c}{A}_7=1.001A_6\\ =1.0001\left({\left(1.0001\right)}^6\left(275000\right)\right)\left[\because A_6={\left(1.0001\right)}^6\left(275000\right)\right]\\ ={\left(1.0001\right)}^7\left(275000\right)\\ =275165.04\end{array}$

Hence, the first eight terms of the sequence $A_n$ to the nearest cent are $275000$, $275027.5$, $275055$, $275027.5$, $275110.02$, $275137.53$, $275165.04$, $275165.04$.

(c)

To determine

The value of $A_n$.

Answer to Problem 1P

The value of $A_n$ is $275000{\left(1.0001\right)}^{n-1}$.

Explanation of Solution

Formula used:

Formula to calculate the nth term of the geometric sequence $A_n$ with first term $A$ and common ratio $R$ is,

$A_n=A{R}^n$ (2)

Given:

Initial amount deposited at the start of 2015 is 275,000 and the rate of interest is $3.65\%\text{per}\text{year}$.

Calculation:

Since, the terms $A_1,A_2,A_3,\mathrm{...}$ in part (b) are in geometric sequence with common ratio $1.0001$ and first term $275000$

Substitute $1.0001$ for $R$ and 275000 for $A$ in equation (2), to evaluate $A_n$.

$A_n=275000{\left(1.001\right)}^n$

Hence, the value of $A_n$ is $275000{\left(1.0001\right)}^{n-1}$.