Chapter 9, Problem 55RE

Solution

To calculate: The balance in the account at the end of 10 years. It is given that Mr. Adabor pays $150 at the end of each month into an account that pays 8% interest compounded monthly.

The balance in account at the end of 10 year is $27,441.91$ dollars.

Given information:

These is given term $a_1=150,r=1+\frac{0.08}{12}\text{and}n=120$ .

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 sequence.

It is given that $a_1=150,r=1+\frac{0.08}{12}\text{and}n=120$ .

Put the value of $a_1=150,r=1+\frac{0.08}{12}\text{and}n=120$ into $s_n=\frac{a_1\left(1-r^n\right)}{1-r}$ .

  $\begin{array}{l}{s}_n=\frac{a_1\left(1-r^n\right)}{1-r}\\ s_{120}=150\cdot \frac{1-{\left(1+\frac{0.08}{12}\right)}^{120}}{1-\left(1+\frac{0.08}{12}\right)}\\ s_{120}=\frac{150\left(1-{1.00666667}^{120}\right)}{-\frac{0.08}{12}}\\ s_{120}\approx 27,441.901\end{array}$