Chapter 8.3, Problem 98E

To determine

To find: the value of $A$ .

Answer to Problem 98E

The value of $A$ is $3806.08.

Explanation of Solution

Given information:

A deposit of $50 is made at the beginning of given month in an account that pays 2% interest, compounded monthly. The balance $A$ in the account at the end of 6 years is given by

  $A=50{\left(1+\frac{0.02}{12}\right)}^1+\mathrm{...}+50{\left(1+\frac{0.02}{12}\right)}^{72}$

Calculation:

The compound interest formula is

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

Now determine the net balance $A$ in the account as shown below:

  $\begin{array}{c}A=50{\left(1+\frac{0.02}{12}\right)}^1+\mathrm{...}+50{\left(1+\frac{0.02}{12}\right)}^{72}\\ a_1=50{\left(1+\frac{0.02}{12}\right)}^1\\ =50\left(1+\frac{0.02}{12}\right)\\ a_2=50{\left(1+\frac{0.02}{12}\right)}^2\\ r=\frac{a_2}{a_1}\\ =50{\left(1+\frac{0.02}{12}\right)}^2\times \frac{1}{50\left(1+\frac{0.02}{12}\right)}\\ =1.001\\ n=72\end{array}$

Now, substitute the value of $a_1,r,n$ in the following equation:

  $\begin{array}{c}{S}_n={\displaystyle \sum _{i=1}^n{a}_1{r}^{i-1}}\\ =a_1\left(\frac{1-r^n}{1-r}\right)\\ A_{72}=50\left(1+\frac{0.02}{12}\right)\times \left(\frac{1-{1.001}^{72}}{1-1.001}\right)\\ =50\left(1+\frac{0.02}{12}\right)\times \left(\frac{1-1.076}{-0.001}\right)\\ =50\times 1.0016\times 76\\ =3806.08\end{array}$

Thus, the value of $A$ is $3806.08.