Suppose you wish to retire at the age of 70 with \( \$ 80,000 \) in savings. Determine your monthly payment into an IRA if the APR is \( 5.5 \% \) compounded monthly and you begin making payments at 25 years old. Round your answer to the nearest cent, if necessary.

To determine the monthly payment into an IRA needed to accumulate \( \$80,000 \) by the age of 70, we can use the future value of an annuity formula. The formula for the future value of an annuity is: \[ FV = P \times \frac{(1 + r)^n - 1}{r} \] Where: - \( FV \) is the future value of the annuity (the amount you want to have at retirement, which is \( \$80,000 \)). - \( P \) is the monthly payment (the value we want to find). - \( r \) is the monthly interest rate (annual interest rate divided by 12). - \( n \) is the total number of payments (number of years until retirement multiplied by 12). ### Step 1: Identify the known values - Future Value, \( FV = 80000 \) - Annual Interest Rate, \( APR = 5.5\% = 0.055 \) - Monthly Interest Rate, \( r = \frac{0.055}{12} \) - Current Age = 25 - Retirement Age = 70 - Total Years until Retirement = \( 70 - 25 = 45 \) - Total Payments, \( n = 45 \times 12 \) ### Step 2: Calculate \( r \) and \( n \) Now, let's calculate \( r \) and \( n \): \[ r = \frac{0.055}{12} \] \[ n = 45 \times 12 \] ### Step 3: Substitute into the formula and solve for \( P \) Rearranging the future value formula to solve for \( P \): \[ P = \frac{FV \times r}{(1 + r)^n - 1} \] Now, let's calculate \( r \), \( n \), and then \( P \). I'll perform these calculations now. Calculate the value by following steps: - step0: Calculate: \(45\times 12\) - step1: Multiply the numbers: \(540\) Calculate or simplify the expression \( 0.055/12 \). Calculate the value by following steps: - step0: Calculate: \(\frac{0.055}{12}\) - step1: Convert the expressions: \(\frac{\frac{11}{200}}{12}\) - step2: Multiply by the reciprocal: \(\frac{11}{200}\times \frac{1}{12}\) - step3: Multiply the fractions: \(\frac{11}{200\times 12}\) - step4: Multiply: \(\frac{11}{2400}\) The calculations yield the following results: - Monthly Interest Rate, \( r \approx 0.00458333 \) - Total Payments, \( n = 540 \) ### Step 4: Substitute into the formula to find \( P \) Now we can substitute \( FV = 80000 \), \( r \approx 0.00458333 \), and \( n = 540 \) into the rearranged formula for \( P \): \[ P = \frac{80000 \times 0.00458333}{(1 + 0.00458333)^{540} - 1} \] Let's calculate \( P \). Calculate the value by following steps: - step0: Calculate: \(\frac{80000\times 0.00458333}{\left(\left(1+0.00458333\right)^{540}-1\right)}\) - step1: Remove the parentheses: \(\frac{80000\times 0.00458333}{\left(1+0.00458333\right)^{540}-1}\) - step2: Add the numbers: \(\frac{80000\times 0.00458333}{1.00458333^{540}-1}\) - step3: Convert the expressions: \(\frac{80000\times 0.00458333}{\left(\frac{100458333}{100000000}\right)^{540}-1}\) - step4: Multiply the numbers: \(\frac{366.6664}{\left(\frac{100458333}{100000000}\right)^{540}-1}\) - step5: Subtract the numbers: \(\frac{366.6664}{\frac{100458333^{540}-100000000^{540}}{100000000^{540}}}\) - step6: Convert the expressions: \(\frac{\frac{458333}{1250}}{\frac{100458333^{540}-100000000^{540}}{100000000^{540}}}\) - step7: Multiply by the reciprocal: \(\frac{458333}{1250}\times \frac{100000000^{540}}{100458333^{540}-100000000^{540}}\) - step8: Rewrite the expression: \(\frac{458333}{1250}\times \frac{1562500^{540}\times 64^{540}}{100458333^{540}-100000000^{540}}\) - step9: Rewrite the expression: \(\frac{458333}{1250}\times \frac{1250^{1080}\times 64^{540}}{100458333^{540}-100000000^{540}}\) - step10: Reduce the numbers: \(458333\times \frac{1250^{1079}\times 64^{540}}{100458333^{540}-100000000^{540}}\) - step11: Multiply: \(\frac{458333\times 1250^{1079}\times 64^{540}}{100458333^{540}-100000000^{540}}\) The calculation gives us: \[ P \approx 33.90 \] ### Conclusion The monthly payment into the IRA needed to accumulate \( \$80,000 \) by the age of 70, with an APR of \( 5.5\% \) compounded monthly, is approximately **\$33.90**.