What APR rate of interest is required to make your initial deposit of $3,776 grow to $6,543 in 13 years, when the interest rate is compounded monthly?

We can use the formula for compound interest to solve for the Annual Percentage Rate (APR). The compound interest formula is:

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

Where:
- $A$ is the final amount ($6,543),
- $P$ is the initial deposit ($3,776),
- $r$ is the annual interest rate (APR, which we want to find),
- $n$ is the number of times the interest is compounded per year (monthly compounding means $n = 12$),
- $t$ is the number of years (13 years).

We are solving for $r$.

### Steps:
1. Plug in the known values: $A = 6543$, $P = 3776$, $n = 12$, and $t = 13$.
2. Rearrange the formula to solve for $r$.

\[
\left( 1 + \frac{r}{12} \right)^{12 \times 13} = \frac{A}{P}
\]

Now, let's calculate the required interest rate.The required Annual Percentage Rate (APR) for your initial deposit of $3,776 to grow to $6,543 in 13 years, with monthly compounding, is approximately **4.24%**.

Would you like more details or have any questions?

Here are five related questions to expand your understanding:
1. How does the compounding frequency affect the final amount in compound interest?
2. What would the final amount be with the same APR but compounded yearly instead of monthly?
3. How does the formula for continuous compounding differ from monthly compounding?
4. How do you calculate the interest rate if you know the future value and initial deposit for other periods?
5. What are the differences between nominal APR and effective annual rate (EAR)?

**Tip:** Monthly compounding grows your money faster than annual compounding due to the frequency of interest calculation!

Learn how to calculate the Annual Percentage Rate (APR) required to grow an initial deposit of $3,776 to $6,543 in 13 years, with monthly compounding. This step-by-step solution covers compound interest concepts, exponential growth, and key mathematical formulas. Suitable for Grades 10-12.

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