Chapter 9, Problem 56RE

Solution

To calculate: The minimum monthly payment at month’s end that must be

made into an account that pays 8% interest compounded monthly if the balance at

the end of 10 years is to be at least $30,000?

It’s practical to make deposits of a fraction of a cent we round up to the closest cent or tenth of a dollar is $\$163.99$ .

Given information:

These is given term $s_n=30000r=0.08,n=12\text{and}t=10$ .

Formula used:

The sum nth term of a geometric sequence with first term and common ratio is given by: $s_n=\frac{a_1\left(1-r^n\right)}{1-r}$

Calculation:

Consider the given term.

It is given that $A=30000r=0.08,n=12\text{and}t=10$ .

Where, $n$ is the number of months per year (12).

t is the number or years.

r is the percent interest.

A is the sum of sequence.

Put the value of $A=30000r=0.08,n=12\text{and}t=10$ into $s_n=a_1\left[\frac{n}{r}\left({\left(1+\frac{r}{n}\right)}^{nt}-1\right)\right]$ .

  $\begin{array}{c}{s}_n=a_1\left[\frac{n}{r}\left({\left(1+\frac{r}{n}\right)}^{nt}-1\right)\right]\\ 30000=a_1\left[\frac{12}{0.08}\left({\left(1+\frac{0.08}{12}\right)}^{120}-1\right)\right]\\ a_1=\frac{30000}{\left[\frac{12}{0.08}\left({\left(1+\frac{0.08}{12}\right)}^{120}-1\right)\right]}\\ a_1=163.9828\end{array}$

So, it’s practical to make deposits of a fraction of a cent we round up to the closest cent or tenth of a dollar is $\$163.99$ .