Chapter 8.3, Problem 102E

(a)

To determine

To find: the balances $A$ after $t$ years when the interest is compound monthly.

Answer to Problem 102E

The required balance is $A=\$38688.25$

Explanation of Solution

Given information:

Consider making monthly deposits of $P$ dollars in a savings account earning an annual interest rate $r$ .

Where

  $P=\$75,r=4\%,t=25\text{years}$

Calculation:

It is stated that a deposit of $P$ dollars is made at the beginning of given month in an account earning an annual interest rate $r$ , compounded monthly. The balance $A$ in the account after $t$ years is given by

  $A=P\left[{\left(1+\frac{r}{12}\right)}^{12t}-1\right]\left(1+\frac{12}{r}\right)$

Now determine the balance $A$ if $P=\$75,r=4\%,t=25\text{years}$ as shown below:

  $\begin{array}{c}A=P\left[{\left(1+\frac{r}{12}\right)}^{12t}-1\right]\left(1+\frac{12}{r}\right)\\ =75\left[{\left(1+\frac{0.04}{12}\right)}^{12\left(25\right)}-1\right]\left(1+\frac{12}{0.04}\right)\\ =75\left[{\left(\frac{12.04}{12}\right)}^{300}-1\right]\left(\frac{12.04}{0.04}\right)\\ =38688.2484405542\\ =\$38688.25\end{array}$

Thus, the required balance is $A=\$38688.25$

(b)

To determine

To find: the balances $A$ after $t$ years when the interest is compound continuously.

Answer to Problem 102E

The required balance is $A=\$38725.81$

Explanation of Solution

Given information:

Consider making monthly deposits of $P$ dollars in a savings account earning an annual interest rate $r$ .

Where

  $P=\$75,r=4\%,t=25\text{years}$

Calculation:

It is stated that a deposit of $P$ dollars is made at the beginning of given month in an account earning an annual interest rate $r$ , compounded monthly. The balance $A$ in the account after $t$ years is given by

  $A=\frac{P{e}^{r/12}\left(e^{rt}-1\right)}{e^{r/12}-1}$

Now determine the balance $A$ if $P=\$75,r=4\%,t=25\text{years}$ as shown below:

  $\begin{array}{c}A=\frac{P{e}^{r/12}\left(e^{rt}-1\right)}{e^{r/12}-1}\\ =\frac{75e^{0.04/12}\left(e^{0.04\left(25\right)}-1\right)}{e^{0.04/12}-1}\\ =\frac{75e^{0.04/12}\left(e^1-1\right)}{e^{0.04/12}-1}\\ =38725.8125064341\\ =\$38725.81\end{array}$

Thus, the required balance is $A=\$38725.81$