How much money should be deposited today in an account that earns 2.5 % compounded monthly so that it will accumulate to $ 10 comma 000 in 3 ​years?
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Part 1
The amount of money that should be deposited is ​$
  
enter your response here.

To solve this problem, we need to use the formula for compound interest:

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- $A$ is the future value (the amount you want to accumulate, which is $10,000),
- $P$ is the present value (the initial amount you need to deposit, which we're solving for),
- $r$ is the annual interest rate (2.5% or 0.025),
- $n$ is the number of times the interest is compounded per year (monthly compounding means $n = 12$),
- $t$ is the number of years (3 years).

Rearranging the formula to solve for $P$:

\[
P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}
\]

Substitute the values:

\[
P = \frac{10,000}{\left(1 + \frac{0.025}{12}\right)^{12 \times 3}}
\]

Let me calculate that for you.The amount of money that should be deposited today is approximately **$9,278.16**.

Would you like more details on the calculation or explanation? Here are some related questions you might find interesting:

1. How would the required deposit change if the interest rate were higher?
2. What is the difference between compound interest and simple interest in this scenario?
3. How much interest will you have earned by the end of the 3 years?
4. What if the interest was compounded quarterly instead of monthly?
5. How would the initial deposit change for a shorter time period, such as 2 years?

**Tip:** When interest is compounded more frequently, the accumulated amount grows faster over time.

How much money should be deposited today in an account that earns 2.5% compounded monthly so that it will accumulate to $10,000 in 3 years?

Discover how to calculate the required deposit to accumulate $10,000 in 3 years with a 2.5% annual interest rate compounded monthly. This guide covers compound interest formulas and provides step-by-step solutions, suitable for students in Grades 10-12.

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