If you deposit $10,000 at 1.85% interest, compounded continuosly, what would your ending balance be after three years?

The formula for continuous compounding is given by:

\[
A = P e^{rt}
\]

where:
- $A$ is the amount of money after time $t$,
- $P$ is the principal (initial amount),
- $r$ is the annual interest rate (as a decimal),
- $t$ is the time in years,
- $e$ is Euler's number ($\approx 2.71828$).

Here:
- $P = 10,000$,
- $r = 1.85\%$ or $0.0185$,
- $t = 3$.

Substituting these values into the formula:

\[
A = 10,000 \times e^{0.0185 \times 3}
\]

Let me calculate that for you.

\[
A = 10,000 \times e^{0.0555}
\] 

Now, calculating further.

\[
A \approx 10,000 \times 1.0571 = 10,571.02
\]

After 3 years, the ending balance would be approximately **$10,571.02**.

Would you like further details or have any questions?

Here are 5 related questions:
1. How does the frequency of compounding affect interest accumulation?
2. How would the result change if the interest was compounded annually instead of continuously?
3. What would be the ending balance after 5 years instead of 3 years?
4. How does continuous compounding compare to simple interest over long periods?
5. What is the impact of increasing the interest rate from 1.85% to 2.5%?

**Tip:** Continuous compounding generally yields slightly higher returns than periodic compounding at the same interest rate.

If you deposit $10,000 at 1.85% interest, compounded continuously, what would your ending balance be after three years?

Find out how to calculate the ending balance of a $10,000 deposit with a 1.85% interest rate, compounded continuously for 3 years. This detailed solution explains key mathematical concepts such as continuous compounding, exponential growth, and Euler’s number, suitable for high school students (Grades 9-12).

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