Chapter 8.3, Problem 103E

(a)

To determine

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

Answer to Problem 103E

The required balance is $A=\$153237.86$

Explanation of Solution

Given information:

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

Where

  $P=\$100,r=5\%,t=40\text{years}$

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 $P=\$100,r=5\%,t=40\text{years}$ as shown below:

  $\begin{array}{c}A=P\left[{\left(1+\frac{r}{12}\right)}^{12t}-1\right]\left(1+\frac{12}{r}\right)\\ =100\left[{\left(1+\frac{0.05}{12}\right)}^{12\left(40\right)}-1\right]\left(1+\frac{12}{0.05}\right)\\ =100\left[{\left(\frac{12.05}{12}\right)}^{480}-1\right]\left(\frac{12.05}{0.05}\right)\\ =153237.857373918\\ =\$153237.86\end{array}$

Thus, the required balance is $A=\$153237.86$

(b)

To determine

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

Answer to Problem 103E

The required balance is $A=\$153657.02$

Explanation of Solution

Given information:

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

Where

  $P=\$100,r=5\%,t=40\text{years}$

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 $P=\$100,r=5\%,t=40\text{years}$ as shown below:

  $\begin{array}{c}A=\frac{P{e}^{r/12}\left(e^{rt}-1\right)}{e^{r/12}-1}\\ =\frac{100e^{0.05/12}\left(e^{0.05\left(40\right)}-1\right)}{e^{0.05/12}-1}\\ =\frac{100e^{0.05/12}\left(e^2-1\right)}{e^{0.05/12}-1}\\ =153657.021021514\\ =\$153657.02\end{array}$

Thus, the required balance is $A=\$153657.02$