Question

Compound Interest If $600 is invested at an interest rate of 2.5% per year, find the amount of the investment at the end of 10 years for the following compounding methods.

(a) Annually

(b) Semiannually

(c) Quarterly

(d) Continuously

Short answer

1. The value of the investment after 10 years is $768.05.
2. The value of the investment after 10 years is $769.22.
3. The value of the investment after 10 years is $769.82.
4. The value of the investment after 10 years is $770.42.

Step-by-step solution

Part a. Step 1. Given.

$$\begin{gathered} P=\mathrm{Principal}=600. \\ r=\mathrm{interest}\mathrm{rate}=2.5\%=0.025. \\ t=\mathrm{number}\mathrm{of}\mathrm{years}=10. \end{gathered}$$

Part a. Step 2. To determine.

We have to find the value of the investment after 10 years for compounded annually.

Part a. Step 3. Calculation.

Let, $A\left(t\right)=\mathrm{Amount}\mathrm{after}t\mathrm{years}$.

$n=\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{compounded}\mathrm{in}\mathrm{a}\mathrm{year}=1$

Because annually means once in a year.

Using compound interest formula: $A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt}$.

Then we plug the values and find $A\left(t\right)$.

$$\begin{gathered} A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt} \\ =600{\left(1+\frac{0.025}{1}\right)}^{1\left(10\right)} \\ =768.050726518 \\ \approx 768.05 \end{gathered}$$

Hence, the value of the investment after 10 years is $768.05.

Part b. Step 1. Given.

$$\begin{gathered} P=\mathrm{Principal}=600. \\ r=\mathrm{interest}\mathrm{rate}=2.5\%=0.025. \\ t=\mathrm{number}\mathrm{of}\mathrm{years}=10. \end{gathered}$$

Part b. Step 2. To determine.

We have to find the value of the investment after 10 years for compounded semiannually.

Part b. Step 3. Calculation.

Let, $A\left(t\right)=\mathrm{Amount}\mathrm{after}t\mathrm{years}$.

$n=\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{compounded}\mathrm{in}\mathrm{a}\mathrm{year}=2$

Because semiannually means twice in a year.

Using compound interest formula: $A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt}$.

Then we plug the values and find $A\left(t\right)$.

$$\begin{gathered} A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt} \\ =600{\left(1+\frac{0.025}{2}\right)}^{2\left(10\right)} \\ =769.222339025 \\ \approx 769.22 \end{gathered}$$

Hence, the value of the investment after 10 years is $769.22.

Part c. Step 1. Given.

$$\begin{gathered} P=\mathrm{Principal}=600. \\ r=\mathrm{interest}\mathrm{rate}=2.5\%=0.025. \\ t=\mathrm{number}\mathrm{of}\mathrm{years}=10. \end{gathered}$$

Part c. Step 2. To determine.

We have to find the value of the investment after 10 years for compounded quarterly.

Part c. Step 3. Calculation.

Let, $A\left(t\right)=\mathrm{Amount}\mathrm{after}t\mathrm{years}$.

$n=\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{compounded}\mathrm{in}\mathrm{a}\mathrm{year}=4$

Because quarterly means 4 times in a year.

Using compound interest formula: $A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt}$.

Then we plug the values and find $A\left(t\right)$.

$$\begin{gathered} A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt} \\ =600{\left(1+\frac{0.025}{4}\right)}^{4\left(10\right)} \\ =769.816092369 \\ \approx 769.82 \end{gathered}$$

Hence, the value of the investment after 10 years is $809.16.

Part d. Step 1. Given.

$$\begin{gathered} P=\mathrm{Principal}=600. \\ r=\mathrm{interest}\mathrm{rate}=2.5\%=0.025. \\ t=\mathrm{number}\mathrm{of}\mathrm{years}=10. \end{gathered}$$

Part d. Step 2. To determine.

We have to find the value of the investment after 10 years for compounded continuously.

Part d. Step 3. Calculation.

Let, $A\left(t\right)=\mathrm{Amount}\mathrm{after}t\mathrm{years}$.

Using continuous compound interest formula: $A\left(t\right)=P{e}^{rt}$.

Then we plug the values and find $A\left(t\right)$.

$$\begin{gathered} A\left(t\right)=P{e}^{rt} \\ =600e^{0.025\left(10\right)} \\ =770.415250013 \\ \approx 770.42 \end{gathered}$$

Hence, the value of the investment after 10 years is $770.42.