The question asks how much money should be deposited today into an account that earns 8% interest compounded semiannually, so that it will accumulate to $10,000 in three years.

This type of problem involves using the formula for compound interest, particularly solving for the present value (the initial deposit, $P$).

The formula for compound interest is:

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

Where:

• $A$ is the amount of money accumulated after $t$ years, including interest (in this case, $10,000).
• $P$ is the principal amount (the initial deposit, which we are solving for).
• $r$ is the annual interest rate (8% or 0.08).
• $n$ is the number of times the interest is compounded per year (since it's compounded semiannually, $n = 2$).
• $t$ is the number of years (3 years).

Step 1: Substitute the known values into the formula.

$10000 = P \left(1 + \frac{0.08}{2}\right)^{2 \times 3}$

$10000 = P \left(1 + 0.04\right)^{6}$

$10000 = P \left(1.04\right)^{6}$

Step 2: Calculate $1.04^6$.

$1.04^6 \approx 1.265319$

Step 3: Solve for $P$.

$10000 = P \times 1.265319$

$P = \frac{10000}{1.265319} \approx 7902.49$

So, the amount that should be deposited today is $7,902.49 (rounded to the nearest cent).

Would you like further details on the steps or concepts? Here are five related questions to explore:

1. How does semiannual compounding affect the growth of interest compared to annual compounding?
2. What would happen if the interest was compounded quarterly instead?
3. How would the amount deposited change if the goal was to accumulate $20,000 in 3 years instead?
4. What is the difference between compound and simple interest in such problems?
5. How would the result change if the interest rate were 6% instead of 8%?

Tip: When dealing with compound interest, the more frequently the interest is compounded, the larger the amount you accumulate over time, given the same interest rate and time duration.