Questions: (a) State when this problem relates to, in ordinary annuity due compounded monthly. After how many months will the account be worth 80,000? (b) Solve this problem (Round your answer up to the nearest month.)

Solution Steps

(a) This problem relates to an ordinary annuity due compounded monthly. To find out after how many months the account will be worth $80,000, we need to use the future value formula for an annuity due. The formula is: \[ FV = P \times \left( \frac{(1 + r)^n - 1}{r} \right) \times (1 + r) \] where \( FV \) is the future value, \( P \) is the payment amount, \( r \) is the monthly interest rate, and \( n \) is the number of periods (months).

(b) To solve this problem, we need to rearrange the formula to solve for \( n \) and then use Python to calculate the value.

We are given the following values:

• Future Value (\( FV \)): \( 80000 \)
• Monthly Payment (\( P \)): \( 1000 \) (assumed for calculation)
• Annual Interest Rate: \( 0.05 \)

From the annual interest rate, we can calculate the monthly interest rate (\( r \)): \[ r = \frac{0.05}{12} = 0.004166666666666667 \]

The future value of an ordinary annuity due is given by the formula: \[ FV = P \times \left( \frac{(1 + r)^n - 1}{r} \right) \times (1 + r) \] To find \( n \), we rearrange the formula: \[ \frac{FV}{P \times (1 + r)} = \frac{(1 + r)^n - 1}{r} \] This leads to: \[ (1 + r)^n = \frac{FV}{P \times (1 + r)} \times r + 1 \]

Taking the logarithm of both sides, we have: \[ n = \frac{\log\left(\frac{FV}{P \times (1 + r)} \times r + 1\right)}{\log(1 + r)} \] Substituting the known values: \[ n = \frac{\log\left(\frac{80000}{1000 \times (1 + 0.004166666666666667)} \times 0.004166666666666667 + 1\right)}{\log(1 + 0.004166666666666667)} \]

After performing the calculations, we find: \[ n \approx 69 \] Rounding up to the nearest month, we conclude that \( n = 69 \).

Final Answer