Chapter 8.3, Problem 104E

(a)

To determine

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

Answer to Problem 104E

The required balance is $A=\$76122.54$

Explanation of Solution

Given information:

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

Where

  $P=\$20,r=6\%,t=50\text{years}$

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 $P=\$20,r=6\%,t=50\text{years}$ as shown below:

  $\begin{array}{c}A=P\left[{\left(1+\frac{r}{12}\right)}^{12t}-1\right]\left(1+\frac{12}{r}\right)\\ =20\left[{\left(1+\frac{0.06}{12}\right)}^{12\left(50\right)}-1\right]\left(1+\frac{12}{0.06}\right)\\ =20\left[{\left(\frac{12.06}{12}\right)}^{600}-1\right]\left(\frac{12.06}{0.06}\right)\\ =76122.5408025588\\ =\$76122.54\end{array}$

Thus, the required balance is $A=\$76122.54$

(b)

To determine

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

Answer to Problem 104E

The required balance is $A=\$76533.16$

Explanation of Solution

Given information:

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

Where

  $P=\$20,r=6\%,t=50\text{years}$

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 $P=\$20,r=6\%,t=50\text{years}$ as shown below:

  $\begin{array}{c}A=\frac{P{e}^{r/12}\left(e^{rt}-1\right)}{e^{r/12}-1}\\ =\frac{20e^{0.06/12}\left(e^{0.06\left(500\right)}-1\right)}{e^{0.06/12}-1}\\ =\frac{20e^{0.06/12}\left(e^3-1\right)}{e^{0.06/12}-1}\\ =76533.1621081168\\ =\$76533.16\end{array}$

Thus, the required balance is $A=\$76533.16$