Question

Compound Interest If $2500 is invested at an interest rate of 2.5% per year, compounded daily, find the value of the investment after the given number of years.

(a) 2 years (b) 3 years (c) 6 years

Short answer

1. The value of the investment after 2 years is $2628.17.
2. The value of the investment after 3 years is $2694.70.
3. The value of the investment after 6 years is $2904.57.

Step-by-step solution

Part a. Step 1. Given.

$P=\mathrm{Principal}=2500$.

$r=\mathrm{interest}\mathrm{rate}=2.5\%=0.025$.

$t=\mathrm{number}\mathrm{of}\mathrm{years}=2$.

$n=\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{compounded}\mathrm{in}\mathrm{a}\mathrm{year}=365$.

Because, interest is compounded daily.

Part a. Step 2. To determine.

We have to find the value of the investment after the given number of years.

Part a. Step-3 – Calculation

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

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

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

$$\begin{gathered} A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt} \\ =2500{\left(1+\frac{0.025}{365}\right)}^{365\left(2\right)} \\ =2628.17324085 \\ \approx 2628.17 \end{gathered}$$

Hence, the value of the investment after 2 years is $2628.17.

Part b. Step 1. Given.

$P=\mathrm{Principal}=2500$.

$r=\mathrm{interest}\mathrm{rate}=2.5\%=0.025$.

$t=\mathrm{number}\mathrm{of}\mathrm{years}=3$.

$n=\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{compounded}\mathrm{in}\mathrm{a}\mathrm{year}=365$.

Because, interest is compounded daily.

Part b. Step 2. To determine.

We have to find the value of the investment after the given number of years.

Part b. Step 3. Calculation.

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

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

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

$$\begin{gathered} A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt} \\ =2500{\left(1+\frac{0.025}{365}\right)}^{365\left(3\right)} \\ =2694.70345619 \\ \approx 2694.70 \end{gathered}$$

Hence, the value of the investment after 3 years is $2694.70.

Part c. Step 1. Given.

$P=\mathrm{Principal}=2500$.

$r=\mathrm{interest}\mathrm{rate}=2.5\%=0.025$.

$t=\mathrm{number}\mathrm{of}\mathrm{years}=6$.

$n=\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{compounded}\mathrm{in}\mathrm{a}\mathrm{year}=365$.

Because, interest is compounded daily.

Part c. Step 2. To determine.

We have to find the value of the investment after the given number of years.

Part c. Step 3. Calculation.

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

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

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

$$\begin{gathered} A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt} \\ =2500{\left(1+\frac{0.025}{365}\right)}^{365\left(6\right)} \\ =2904.57068672 \\ \approx 2904.57 \end{gathered}$$

Hence, the value of the investment after 6 years is $2904.57.