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

To calculate the ending balance of an investment with continuously compounded interest, we use the formula:

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

Where:
- $A$ is the ending balance (the amount after interest is applied),
- $P$ is the principal (initial amount),
- $r$ is the annual interest rate (as a decimal),
- $t$ is the time in years,
- $e$ is the base of the natural logarithm (approximately 2.71828).

In this case:
- $P = 10,000$,
- $r = 1.85\% = 0.0185$,
- $t = 3$ years.

Substitute these values into the formula:

\[
A = 10,000 \cdot e^{(0.0185 \cdot 3)}
\]

Let me calculate that.The ending balance after 3 years, with continuously compounded interest at a rate of 1.85%, would be approximately \$10,570.69.

Would you like further details on this calculation, or do you have any other questions?

Here are some related questions:
1. How does continuously compounded interest differ from compound interest compounded monthly?
2. What would the balance be after 5 years instead of 3 years?
3. What is the formula for compound interest if it’s compounded annually?
4. How would the ending balance change if the interest rate increased to 2.5%?
5. How does the time period affect the growth of the investment with continuous compounding?

**Tip:** Continuous compounding results in slightly higher returns compared to standard compound interest, especially over longer periods.

Learn how to calculate the ending balance of a $10,000 investment at 1.85% interest, compounded continuously, over 3 years. This problem uses the formula A = P * e^(rt) and explores the concept of continuous compounding for exponential growth. Suitable for high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation