If you need $17,000 in 11 years, and can invest in an account with 8.75% monthly compounded interest, how much would you need to deposit today?

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

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

Where:
- $A$ is the future value of the investment/loan, which is $17,000.
- $P$ is the principal amount (the initial deposit we want to find).
- $r$ is the annual interest rate (8.75% or 0.0875 as a decimal).
- $n$ is the number of times the interest is compounded per year (monthly, so $n = 12$).
- $t$ is the time the money is invested for (11 years).

We need to solve for $P$, the initial deposit. Rearranging the formula:

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

Substituting the known values:

\[
P = \frac{17,000}{\left(1 + \frac{0.0875}{12}\right)^{12 \times 11}}
\]

Let me calculate this.The initial deposit you would need to make today is approximately $6,515.64 in order to reach $17,000 in 11 years with an 8.75% interest rate compounded monthly.

Would you like any further details or clarifications? Here are some related questions:

1. How does the frequency of compounding affect the final amount?
2. What happens if the interest rate changes midway through the investment?
3. How does the result change if the investment period is longer or shorter?
4. What is the difference between monthly and yearly compounded interest?
5. How can this formula be applied to different interest rates or periods?

**Tip:** The more frequently interest is compounded, the more interest you will earn over time due to the effect of compounding on a more frequent basis.

Find out how much you need to deposit today to accumulate $17,000 in 11 years with an 8.75% monthly compounded interest rate. This solution includes the compound interest formula and explains key concepts like exponential growth and compounding frequency.

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