1. **State the problem:** We want to find the number of monthly deposits of 85 needed to reach a balance of 18700 in an account earning 3.1% per annum compounded quarterly. 2. **Identify the formula:** Since deposits are monthly but interest compounds quarterly, we use the future value of an annuity formula adjusted for compounding periods: $$FV = P \times \frac{(1 + i)^n - 1}{i}$$ where: - $FV$ is the future value (18700), - $P$ is the monthly deposit (85), - $i$ is the interest rate per deposit period, - $n$ is the number of deposits. 3. **Calculate the interest rate per deposit period:** - Annual nominal rate = 3.1% = 0.031 - Compounded quarterly means 4 compounding periods per year. - Interest rate per quarter = $\frac{0.031}{4} = 0.00775$ Since deposits are monthly but compounding is quarterly, we find the effective monthly interest rate: $$i = (1 + 0.00775)^{\frac{1}{3}} - 1$$ Calculate: $$i = (1.00775)^{0.3333} - 1 \approx 0.00258$$ 4. **Set up the equation:** $$18700 = 85 \times \frac{(1 + 0.00258)^n - 1}{0.00258}$$ 5. **Isolate $(1 + i)^n$:** $$\frac{18700 \times 0.00258}{85} = (1.00258)^n - 1$$ Calculate left side: $$\frac{18700 \times 0.00258}{85} = \frac{48.246}{85} \approx 0.5676$$ So: $$(1.00258)^n - 1 = 0.5676$$ 6. **Add 1 to both sides:** $$(1.00258)^n = 1.5676$$ 7. **Take natural logarithm of both sides:** $$n \ln(1.00258) = \ln(1.5676)$$ 8. **Solve for $n$:** $$n = \frac{\ln(1.5676)}{\ln(1.00258)}$$ Calculate: $$\ln(1.5676) \approx 0.448$$ $$\ln(1.00258) \approx 0.00258$$ So: $$n = \frac{0.448}{0.00258} \approx 173.6$$ 9. **Interpretation:** Since $n$ must be a whole number of deposits, round up: $$n = 174$$ **Final answer:** The number of deposits needed is **174**.