Chapter 12.4, Problem 10MCQ

To determine

To find: the value of the investment at the end of the year.

Answer to Problem 10MCQ

  $\$2154.57$ .

Explanation of Solution

Given:

The interest is compound each quarter so, $n=4$ and the interest rate per period is:

  $\frac{12\%}{4}=3\%$ .

Concept used:

If there is common ratio then the series is geometric series.

Common ratio is:

  $\begin{array}{l}r=\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=k.\\ t_n=a{r}^{n-1}.\\ s_n=\frac{r\left(1-r^n\right)}{1-r}.\end{array}$

Calculation:

According to the given that

The interest is compound each quarter so, $n=4$ and the interest rate per period is:

  $\frac{12\%}{4}=3\%$ .

The common ratio $r$ for the geometric series is then:

  $\begin{array}{l}r=1+\mathrm{0.03.}\\ \Rightarrow \mathrm{1.03.}\end{array}$

The first term $a_1$ in this series is the account balance at the end of the first quarter.

Therefore, the balance of the account at the end of the first quarter is

  $\begin{array}{l}{a}_1=\left(\$500\right)\left(1.03\right).\\ \Rightarrow \$515.\end{array}$

Apply for the formula for the sum of the geometric series.

  $s_n=\frac{a_1-a_1{r}^n}{1-r}$ .

For the end of the years, $n=4$ .

Total savings of him is:

  $\begin{array}{l}{S}_n=\frac{\left(\$515\right)-\left(\$515\right){\left(1.03\right)}^4}{1-1.03}.\\ \Rightarrow \$\mathrm{2154.57.}\end{array}$

Hence, $\$2154.57$ .