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

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

Where:

• $A$ is the final amount (in this case, $6,200),
• $P$ is the initial investment (which we need to find),
• $r$ is the annual interest rate (4.2%, or 0.042),
• $n$ is the number of times the interest is compounded per year (quarterly, so $n = 4$),
• $t$ is the number of years (19 years).

Step 1: Plug in the known values

$6,200 = P \left(1 + \frac{0.042}{4}\right)^{4 \times 19}$

Step 2: Simplify the equation

First, simplify the term inside the parentheses:

$\frac{0.042}{4} = 0.0105$

Now, the formula becomes:

$6,200 = P \left(1 + 0.0105\right)^{76}$

$6,200 = P \left(1.0105\right)^{76}$

Step 3: Calculate the exponentiation

Next, calculate $\left(1.0105\right)^{76}$:

$\left(1.0105\right)^{76} \approx 2.337$

Step 4: Solve for $P$

Now, solve for $P$:

$6,200 = P \times 2.337$

$P = \frac{6,200}{2.337} \approx 2,653.04$

Step 5: Round to the nearest dollar

$P \approx 2,653$

So, Adam would need to invest $2,653 to reach $6,200 in 19 years with an interest rate of 4.2% compounded quarterly.

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Would you like more details on how the compound interest works? Or do you have any other questions?

Here are some related questions for you to explore further:

1. How does changing the compounding frequency (e.g., monthly or annually) affect the investment amount?
2. What would the future value be if Adam invested $2,653 at a rate of 5% compounded quarterly for 19 years?
3. How would the investment change if the interest rate were lower, say 3% instead of 4.2%?
4. How do you calculate compound interest when the time period is in months instead of years?
5. What are some real-world examples where compound interest is used?

Tip: The more frequently interest is compounded, the more interest is accumulated over time.