Let's break down the problem step by step: ### **Given:** - Goal (FV): \$10,000 - Annual deposit (PMT): \$1,150 (except possibly the last deposit) - Interest rate (r): 8% per year - First payment is made 1 year from today (ordinary annuity) - Last deposit may be less than \$1,150 to round out to \$10,000 --- ## **Step 1: Formula for Future Value of an Ordinary Annuity** The future value of an ordinary annuity is: \[ FV = PMT \times \frac{(1 + r)^n - 1}{r} \] Where: - \(FV\) = Future Value (\$10,000) - \(PMT\) = Payment per period (\$1,150) - \(r\) = Interest rate per period (0.08) - \(n\) = Number of periods (years, to be found) --- ## **Step 2: Plug in the values and solve for \(n\)** \[ 10,000 = 1,150 \times \frac{(1 + 0.08)^n - 1}{0.08} \] \[ 10,000 = 1,150 \times \frac{(1.08)^n - 1}{0.08} \] \[ \frac{10,000}{1,150} = \frac{(1.08)^n - 1}{0.08} \] \[ 8.69565 = \frac{(1.08)^n - 1}{0.08} \] \[ 8.69565 \times 0.08 = (1.08)^n - 1 \] \[ 0.695652 = (1.08)^n - 1 \] \[ 0.695652 + 1 = (1.08)^n \] \[ 1.695652 = (1.08)^n \] --- ## **Step 3: Solve for \(n\) using logarithms** \[ (1.08)^n = 1.695652 \] \[ n = \frac{\ln(1.695652)}{\ln(1.08)} \] \[ n = \frac{0.52887}{0.07696} \] \[ n = 6.87 \] Round **up to the nearest whole number**: **7 years** --- ## **Step 4: Find the size of the last deposit** We need to find out how much will be required in the 7th year to reach exactly \$10,000. First, calculate the accumulated amount after 6 full deposits: \[ FV_6 = 1,150 \times \frac{(1.08)^6 - 1}{0.08} \] \[ (1.08)^6 = 1.586874 \] \[ FV_6 = 1,150 \times \frac{1.586874 - 1}{0.08} \] \[ FV_6 = 1,150 \times \frac{0.586874}{0.08} \] \[ FV_6 = 1,150 \times 7.335925 \] \[ FV_6 = 8,436.31 \] Now, this amount will grow for one more year at 8% interest: \[ FV_{6, \text{after 1 year}} = 8,436.31 \times 1.08 = 9,110.22 \] Now, the last deposit (\(X\)) will be made at the end of year 7, so it does **not** earn interest. \[ \text{Total FV after 7 years} = 9,110.22 + X = 10,000 \] \[ X = 10,000 - 9,110.22 = 889.78 \] --- ## **Step 5: Final Answers** ### (a) **Number of years:** **7 years** ### (b) **Last deposit amount:** **\$889.78** --- ### **Summary Table** | Step | Value | |--------------------------|------------| | Number of years (n) | 7 years | | Last deposit (rounded) | \$889.78 | --- **Final Answers:** - **Years:** 7 - **Last deposit:** \$889.78

Let's carefully work through the problem step by step, providing detailed calculations and explanations. --- ### **Problem Recap:** - You want to save **\$10,000**. - You will deposit **\$1,150** annually, starting **1 year from today**. - The bank pays **8% interest** annually. - The last deposit may be less than \$1,150 if needed to reach exactly \$10,000. --- ### **Step 1: Understand the type of problem** This is a future value (FV) of an **ordinary annuity** problem, where deposits are made at the end of each period, and interest accrues annually. **Given:** - \(FV = \$10,000\) - \(PMT = \$1,150\) - \(r = 8\% = 0.08\) - Payments are made **annually**, starting 1 year from today. --- ### **Step 2: Use the FV of an ordinary annuity formula** \[ FV = PMT \times \frac{(1 + r)^n - 1}{r} \] We need to find **n**, the number of years. --- ### **Step 3: Set up the equation** \[ 10,000 = 1,150 \times \frac{(1.08)^n - 1}{0.08} \] Divide both sides by 1,150: \[ \frac{10,000}{1,150} = \frac{(1.08)^n - 1}{0.08} \] Calculate: \[ 8.69565 = \frac{(1.08)^n - 1}{0.08} \] Multiply both sides by 0.08: \[ 8.69565 \times 0.08 = (1.08)^n - 1 \] \[ 0.69565 = (1.08)^n - 1 \] Add 1 to both sides: \[ (1.08)^n = 1 + 0.69565 = 1.69565 \] --- ### **Step 4: Solve for \(n\) using logarithms** \[ n = \frac{\ln(1.69565)}{\ln(1.08)} \] Calculate numerator: \[ \ln(1.69565) \approx 0.527 \] Calculate denominator: \[ \ln(1.08) \approx 0.077 \] Now: \[ n = \frac{0.527}{0.077} \approx 6.84 \] Since fractional years are not practical for the number of deposits, **round up**: \[ \boxed{ n = 7 \text{ years} } \] --- ### **Step 5: Calculate the total amount after 6 years** To determine the size of the last deposit, first compute the accumulated amount after 6 full deposits, just before the 7th deposit. **Calculate FV after 6 deposits:** \[ FV_6 = 1,150 \times \frac{(1.08)^6 - 1}{0.08} \] Calculate \((1.08)^6\): \[ (1.08)^6 \approx 1.586874 \] Compute numerator: \[ 1.586874 - 1 = 0.586874 \] Compute FV: \[ FV_6 = 1,150 \times \frac{0.586874}{0.08} = 1,150 \times 7.335925 \approx 8,436.31 \] --- ### **Step 6: Grow the amount after 6 years to the end of year 7** The amount after 6 deposits will grow for one more year at 8%: \[ FV_{after\, 6\, years} = 8,436.31 \times 1.08 \approx 9,110.22 \] ### **Step 7: Determine last deposit amount** To reach **\$10,000** at the end of year 7: \[ \text{Last deposit} = 10,000 - 9,110.22 = \$889.78 \] This is the size of the **final deposit** needed. --- ### **### Final Answers:** **a) Number of years needed:** \[ \boxed{ \textbf{7 years} } \] **b) Size of the last deposit:** \[ \boxed{ \$889.78 } \] --- ### **Summary:** | Step | Calculation | Result | |---------|-----------------------|---------| | Number of years | \(\frac{\ln(1.69565)}{\ln(1.08)} \approx 6.84\) | **7 years** (rounded up) | | Last deposit | \(10,000 - 8,436.31 \times 1.08 \approx 889.78\) | **\$889.78** | This detailed calculation confirms it will take **7 years** to reach the \$10,000 goal, with the final deposit being approximately **\$889.78**.