Chapter 3.1, Problem 68E

To determine

Find the balance A for $\$2500$ for different value of n .

Answer to Problem 68E

The value for,

  $\begin{array}{l}n=1\to 8155\\ n=2\to 8225\\ n=4\to 8262.50\\ n=12\to 8287.50\\ n=365\to 8288.75\\ n=continous\to 8300.7\end{array}$

Explanation of Solution

Given:

It is given in the question that Principal amount is $\$2500$ at the rate of $3\%$ and the time is $40$ .

Concept Used:

In this,use the concept of the compound interest and here use the formula of $A=P{(1+\frac{r}{n})}^{nt}$ .

Calculation:

In this, P = $\$2500$ , r = $3\%$ = $0.03$ ,t = $40$ years.

Now , put the above values on $A=P{(1+\frac{r}{n})}^{nt}$ by taking $n=1$ ,

  $\begin{array}{}$ A=P{(1+\frac{r}{n})}^{nt}$\end{array}$

  $A=2500{(1+\frac{0.03}{1})}^{1\times 40}$

  $A=2500{(1.03)}^{40}$

  $\begin{array}{l}A=2500\times 3.262\\ A=8155\end{array}$

Now,put $n=2$ ,

  $\begin{array}{}$ A=P{(1+\frac{r}{n})}^{nt}$\end{array}$

  $A=2500{(1+\frac{0.03}{2})}^{2\times 40}$

  $A=2500{(1.015)}^{80}$

  $\begin{array}{l}A=2500\times 3.290\\ A=8225\end{array}$

Now,put $n=4$ ,

  $\begin{array}{}$ A=P{(1+\frac{r}{n})}^{nt}$\end{array}$

  $A=2500{(1+\frac{0.03}{4})}^{4\times 40}$

  $A=2500{(1.0075)}^{160}$

  $\begin{array}{l}A=2500\times 3.305\\ A=8262.50\end{array}$

Now,put $n=12$ ,

  $\begin{array}{}$ A=P{(1+\frac{r}{n})}^{nt}$\end{array}$

  $A=2500{(1+\frac{0.03}{12})}^{12\times 40}$

  $A=2500{(1.0025)}^{480}$

  $\begin{array}{l}A=2500\times 3.315\\ A=8287.50\end{array}$

Now,put $n=365$ ,

  $\begin{array}{}$ A=P{(1+\frac{r}{n})}^{nt}$\end{array}$

  $A=2500{(1+\frac{0.03}{365})}^{365\times 40}$

  $A=2500{(1.0000821)}^{14600}$

  $\begin{array}{l}A=2500\times 3.3155\\ A=8288.75\end{array}$

And Now put n = continous

  $\begin{array}{l}A=P{e}^{rt}\\ A=2500e^{0.03\times 40}\\ A=2500e^{1.2}\\ A=2500\times 3.320=8300.7\end{array}$

Conclusion:

The value for,

  $\begin{array}{l}n=1\to 8155\\ n=2\to 8225\\ n=4\to 8262.50\\ n=12\to 8287.50\\ n=365\to 8288.75\\ n=continous\to 8300.7\end{array}$