Chapter 3.1, Problem 67E

To determine

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

Answer to Problem 67E

The value for,

  $\begin{array}{l}n=1\to 5477.50\\ n=2\to 5520.099\\ n=4\to 5541.75\\ n=12\to 5556\\ n=365\to 5556.15\\ n=continous\to 5563.75\end{array}$

Explanation of Solution

Given:

It is given in the question that Principal amount is $\$2500$ at the rate of $4\%$ and the time is $20$ .

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 = $4\%$ = $0.04$ ,t = $20$ 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}$\end{array}$

  $A=2500{(1+\frac{0.04}{1})}^{1\times 20}$

  $A=2500{(1.04)}^{20}$

  $\begin{array}{l}A=2500\times 2.191\\ A=5477.50\end{array}$

Now,put $n=2$ ,

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

  $A=2500{(1+\frac{0.04}{2})}^{2\times 20}$

  $A=2500{(1.02)}^{40}$

  $\begin{array}{l}A=2500\times 2.208\\ A=5520.099\end{array}$

Now,put $n=4$ ,

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

  $A=2500{\left(1+\frac{0.04}{4}\right)}^{4\times 20}$

  $A=2500{(1.01)}^{80}$

  $A=2500\times 2.2167$

  $A=5541.75$

Now,put $n=12$ ,

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

  $A=2500{\left(1+\frac{0.04}{12}\right)}^{12\times 20}$

  $A=2500{(1.003333)}^{240}$

  $A=2500\times 2.2224$

  $A=5556$

Now,put $n=365$ ,

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

  $A=2500{\left(1+\frac{0.04}{365}\right)}^{365\times 20}$

  $A=2500{(1.000109)}^{7300}$

  $A=2500\times 2.22246$

  $A=5556.15$

And Now put n = continous

  $\begin{array}{l}A=P{e}^{rt}\\ A=2500e^{0.04\times 20}\\ A=2500e^{0.8}\\ A=2500\times 2.2255=5563.75\end{array}$

Conclusion:

The value for,

  $\begin{array}{l}n=1\to 5477.50\\ n=2\to 5520.099\\ n=4\to 5541.75\\ n=12\to 5556\\ n=365\to 5556.15\\ n=continous\to 5563.75\end{array}$