Chapter 3.1, Problem 52E

To determine

To Calculate: The balance when $\$1000$ is invested.

Answer to Problem 52E

n	1	2	4	12	365	Continuous
A	$\$10285.72$	$\$10640.89$	$\$10828.46$	$\$10957.45$	$\$\text{11021}\text{.00}$	$\$\text{11023}\text{.18}$

Explanation of Solution

Given: Principle amount, $P=\$1000$

Rate of interest, $r=6\%$

Number of years for which investment is done, $t=40\text{years}$

Number of times, interest is compounded,

  $\begin{array}{l}{n}_1=1\\ n_2=2\\ n_3=4\\ n_4=12\\ n_5=365\\ n_6=\text{continuous}\end{array}$

Formula Used:

Amount can be obtained by:

  $A=P{\left(1+\frac{r}{n}\right)}^{nt}$

If the interest is compounded continuously, then the amount can be obtained by:

  $A=P{e}^{rt}$

Calculations:

For $n=1$

  $\begin{array}{c}A=\$1000{\left(1+\frac{6}{1\times 100}\right)}^{1\times 40}\\ =\$10285.72\end{array}$

For $n=2$

  $\begin{array}{c}A=\$1000{\left(1+\frac{6}{2\times 100}\right)}^{2\times 40}\\ =\$10640.89\end{array}$

For $n=4$

  $\begin{array}{c}A=\$1000{\left(1+\frac{6}{4\times 100}\right)}^{4\times 40}\\ =\$10828.46\end{array}$

For $n=12$

  $\begin{array}{c}A=\$1000{\left(1+\frac{6}{12\times 100}\right)}^{12\times 40}\\ =\$10957.45\end{array}$

For $n=365$

  $\begin{array}{c}A=\$1000{\left(1+\frac{6}{365\times 100}\right)}^{365\times 40}\\ =\$\text{11021}\text{.00}\end{array}$

For $n=\text{continuously}$

  $\begin{array}{c}A=P{e}^{rt}\\ =\$1500e^{6\times 40}\\ =\$\text{11023}\text{.18}\end{array}$