1. **State the problem:** Paul wants to accumulate at least 55000 by depositing 2000 at the end of every month into a fund with 6.25% annual interest compounded monthly. We need to find how many deposits (n) are required and the time in years and months. 2. **Formula for future value of an ordinary annuity:** $$FV = P \times \frac{(1 + r)^n - 1}{r}$$ where $P$ is the monthly deposit, $r$ is the monthly interest rate, and $n$ is the number of deposits. 3. **Identify values:** - $FV = 55000$ - $P = 2000$ - Annual interest rate = 6.25% so monthly rate $r = \frac{6.25}{100 \times 12} = 0.0052083$ 4. **Set up equation:** $$55000 = 2000 \times \frac{(1 + 0.0052083)^n - 1}{0.0052083}$$ 5. **Isolate the term with $n$:** $$\frac{55000 \times 0.0052083}{2000} = (1.0052083)^n - 1$$ Calculate left side: $$\frac{55000 \times 0.0052083}{2000} = 0.143229$$ 6. **Add 1 to both sides:** $$(1.0052083)^n = 1 + 0.143229 = 1.143229$$ 7. **Take natural logarithm of both sides:** $$\ln((1.0052083)^n) = \ln(1.143229)$$ $$n \ln(1.0052083) = 0.1337$$ 8. **Solve for $n$:** $$n = \frac{0.1337}{\ln(1.0052083)}$$ Calculate denominator: $$\ln(1.0052083) = 0.005195$$ So, $$n = \frac{0.1337}{0.005195} = 25.74$$ 9. **Round up to next whole deposit:** $$n = 26$$ deposits 10. **Convert deposits to time:** Each deposit is monthly, so 26 months. 11. **Convert months to years and months:** $$26 \text{ months} = 2 \text{ years and } 2 \text{ months}$$ **Final answers:** - a. Number of deposits needed: 26 - b. Time to reach goal: 2 years and 2 months