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

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

where:

• $A$ = final amount
• $P$ = principal (initial deposit)
• $r$ = annual interest rate (as a decimal)
• $n$ = number of times interest is compounded per year
• $t$ = number of years

Step 1: Calculate the amount after 3 years (first period)

• Initial deposit: $P = 4000$
• Interest rate: $r = 8\% = 0.08$
• Compounded semiannually ($n = 2$)
• Time: $t = 3$

$A_1 = 4000 \left(1 + \frac{0.08}{2}\right)^{2 \times 3}$

$A_1 = 4000 \left(1.04\right)^6$

$A_1 \approx 4000 \times 1.2653$

$A_1 \approx 5061.20$

Step 2: Calculate the amount after the next 3 years (second period)

• New principal: $P = A_1 \approx 5061.20$
• New interest rate: $r = 8.08\% = 0.0808$
• Compounded quarterly ($n = 4$)
• Time: $t = 3$

$A_2 = 5061.20 \left(1 + \frac{0.0808}{4}\right)^{4 \times 3}$

$A_2 = 5061.20 \left(1.0202\right)^{12}$

$A_2 \approx 5061.20 \times 1.2685$

$A_2 \approx 6421$

Final Answer:

$\mathbf{\$6421}$

Would you like me to clarify any step or explain further?

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Related Questions:

1. How does compounding frequency affect the final amount in an account?
2. What happens if interest is compounded continuously?
3. How would the answer change if the interest rate increased after 4 years instead of 3?
4. How do you convert an annual interest rate to a per-period interest rate?
5. What is the difference between simple interest and compound interest?

Tip: The more frequently interest is compounded, the higher the final amount will be!