## Problem Restatement - **Goal:** Have $50,000 in 5 years for a down payment. - **Initial savings:** $20,000 (today) - **Quarterly contributions:** $500 at the end of each quarter (every 3 months) - **Interest:** Compounded monthly at nominal annual rate \( r \) - **Find:** The nominal annual rate \( r \) (compounded monthly) that allows you to reach $50,000 in 5 years. --- ## Step 1: Timeline and Parameters - **Time horizon:** 5 years = 60 months - **Number of quarters:** \( 5 \text{ years} \times 4 = 20 \) quarters - **Monthly compounding:** Interest is credited monthly. --- ## Step 2: Formula for Future Value We need the future value (FV) of both: 1. The **lump sum** $20,000 deposited today. 2. The **series of quarterly deposits** of $500 each. Let: - \( r \) = nominal annual rate (compounded monthly) - \( i = \frac{r}{12} \) = monthly rate - \( n = 60 \) = number of months --- ### 2.1: Future Value of Lump Sum \[ FV_{\text{lump}} = PV \times (1+i)^n \] where \( PV = 20,000 \). --- ### 2.2: Future Value of Quarterly Deposits Each $500 is deposited at the end of every quarter (every 3 months). Each deposit earns interest for a different number of months until the end of year 5. Each $500 deposit is made at months: 3, 6, 9, ..., 60 (i.e., months \( 3k \); \( k=1 \) to \( 20 \)). The FV of the \( k \)th deposit (made at month \( m = 3k \)): \[ FV_k = 500 \times (1+i)^{60-m} \] Total FV of all contributions: \[ FV_{\text{series}} = \sum_{k=1}^{20} 500 \times (1+i)^{60-3k} \] --- ## Step 3: Set up the Equation Total FV after 5 years: \[ FV_{\text{total}} = FV_{\text{lump}} + FV_{\text{series}} = 50,000 \] \[ 20,000 \times (1+i)^{60} + 500 \sum_{k=1}^{20} (1+i)^{60-3k} = 50,000 \] --- ## Step 4: Simplify the Series Notice that \( (1+i)^{60-3k} = (1+i)^{60} \times [(1+i)^{-3}]^k \). Let \( A = (1+i)^{60} \), \( q = (1+i)^{-3} \). So: \[ \sum_{k=1}^{20} (1+i)^{60-3k} = (1+i)^{60} \sum_{k=1}^{20} [(1+i)^{-3}]^k = A \sum_{k=1}^{20} q^k \] Sum of a geometric series: \[ \sum_{k=1}^{N} q^k = q \frac{1 - q^{N}}{1 - q} \] Here, \( N = 20 \). --- ## Step 5: Substitute Back \[ 20,000A + 500A \cdot \left[ q \frac{1 - q^{20}}{1 - q} \right] = 50,000 \] \[ A \left[ 20,000 + 500 \cdot \frac{q(1 - q^{20})}{1 - q} \right] = 50,000 \] --- ## Step 6: Express in Terms of \( i \) Recall: - \( A = (1+i)^{60} \) - \( q = (1+i)^{-3} \) \[ (1+i)^{60} \left[ 20,000 + 500 \cdot \frac{(1+i)^{-3} (1 - (1+i)^{-60})}{1 - (1+i)^{-3}} \right] = 50,000 \] --- ## Step 7: Solve Numerically Let’s use trial-and-error or a financial calculator (or spreadsheet) to solve for \( i \). Let’s denote \( i = \frac{r}{12} \). We’ll try different values for \( i \) to converge to the required \( r \). --- ### Try \( i = 0.005 \) (i.e., 6% nominal, monthly compounding) - \( (1+i)^{60} = (1.005)^{60} \approx 1.34885 \) - \( (1+i)^{-3} = (1.005)^{-3} \approx 0.9851 \) - \( (1+i)^{-60} \approx \frac{1}{1.34885} \approx 0.74145 \) Now: \[ \text{Denominator:} \ 1 - (1+i)^{-3} = 1 - 0.9851 = 0.0149 \] \[ \text{Numerator:} \ (1+i)^{-3}(1 - (1+i)^{-60}) = 0.9851 \times (1 - 0.74145) = 0.9851 \times 0.25855 \approx 0.2547 \] \[ \frac{\text{Numerator}}{\text{Denominator}} = \frac{0.2547}{0.0149} \approx 17.10 \] \[ \text{Bracket:} \ 20,000 + 500 \times 17.10 = 20,000 + 8,550 = 28,550 \] \[ FV = (1.34885) \times 28,550 \approx 38,539 \] This is **less than $50,000**, so try a higher rate. --- ### Try \( i = 0.01 \) (i.e., 12% nominal) - \( (1.01)^{60} \approx 1.8167 \) - \( (1.01)^{-3} \approx 0.9706 \) - \( (1.01)^{-60} \approx 0.5503 \) \[ 1 - 0.9706 = 0.0294 \] \[ 0.9706 \times (1 - 0.5503) = 0.9706 \times 0.4497 \approx 0.4360 \] \[ 0.4360 / 0.0294 \approx 14.83 \] \[ 20,000 + 500 \times 14.83 = 20,000 + 7,415 = 27,415 \] \[ 1.8167 \times 27,415 \approx 49,785 \] Close, but still slightly less. Try \( i = 0.0102 \) (12.24% nominal): --- ### Try \( i = 0.01025 \) (12.3% nominal): - \( (1.01025)^{60} \approx e^{60 \times \ln(1.01025)} \approx e^{60 \times 0.0101976} \approx e^{0.61186} \approx 1.8434 \) - \( (1.01025)^{-3} \approx 1/(1.01025)^3 \approx 1/1.031 \approx 0.9699 \) - \( (1.01025)^{-60} \approx 1/1.8434 \approx 0.5423 \) - \( 1 - 0.9699 = 0.0301 \) - \( 0.9699 \times (1 - 0.5423) = 0.9699 \times 0.4577 \approx 0.4441 \) - \( 0.4441 / 0.0301 \approx 14.75 \) - \( 20,000 + 500 \times 14.75 = 20,000 + 7,375 = 27,375 \) - \( 1.8434 \times 27,375 \approx 50,489 \) Now, **this is slightly more than $50,000**. --- ### Interpolating Let’s interpolate between \( i = 0.01 \) (FV = 49,785) and \( i = 0.01025 \) (FV = 50,489): \[ r = 12\% + \frac{(50,000 - 49,785)}{(50,489 - 49,785)} \times (12.3\% - 12\%) \] \[ = 12\% + \frac{215}{704} \times 0.3\% \] \[ = 12\% + 0.0917\% \] \[ = 12.09\% \text{ nominal annual rate} \] --- ## **Final Answer** ### **You must earn a nominal annual rate of approximately \(\boxed{12.1\%}\) compounded monthly to reach your $50,000 goal in 5 years.** --- ### **Summary Table** | i (monthly rate) | r (nominal) | FV | |------------------|-------------|------------| | 0.005 | 6.00% | $38,539 | | 0.0100 | 12.00% | $49,785 | | 0.01025 | 12.3% | $50,489 | Interpolated solution: **12.1% nominal annual, compounded monthly**.