To solve this problem, we need to determine the size of the equal payments that will settle the loan of $6,000, given that the payments are due in 2 months and 4 months, and the interest rate is 3.50% per annum. We will use the loan date as the focal date.

Step 1: Convert the annual interest rate to a monthly interest rate.
The annual interest rate is 3.50%. To find the monthly interest rate, we divide by 12:
$$ \text{Monthly interest rate} = \frac{3.50\%}{12} = \frac{0.035}{12} \approx 0.0029167 $$

Step 2: Calculate the present value of the payments.
Let $ P $ be the size of each equal payment. The present value of the first payment (due in 2 months) and the second payment (due in 4 months) must equal the loan amount of $6,000.

The present value of the first payment (due in 2 months) is:
$$ PV_1 = \frac{P}{(1 + 0.0029167)^2} $$

The present value of the second payment (due in 4 months) is:
$$ PV_2 = \frac{P}{(1 + 0.0029167)^4} $$

Step 3: Set up the equation for the total present value.
The total present value of the payments must equal the loan amount:
$$ PV_1 + PV_2 = 6000 $$
$$ \frac{P}{(1 + 0.0029167)^2} + \frac{P}{(1 + 0.0029167)^4} = 6000 $$

Step 4: Simplify the equation.
First, calculate the denominators:
$$ (1 + 0.0029167)^2 \approx 1.00584 $$
$$ (1 + 0.0029167)^4 \approx 1.01171 $$

Now, substitute these values into the equation:
$$ \frac{P}{1.00584} + \frac{P}{1.01171} = 6000 $$

Step 5: Combine the terms.
$$ \frac{P}{1.00584} + \frac{P}{1.01171} = 6000 $$
$$ P \left( \frac{1}{1.00584} + \frac{1}{1.01171} \right) = 6000 $$

Step 6: Calculate the sum of the fractions.
$$ \frac{1}{1.00584} \approx 0.99419 $$
$$ \frac{1}{1.01171} \approx 0.98842 $$
$$ 0.99419 + 0.98842 \approx 1.98261 $$

Step 7: Solve for $ P $.
$$ P \times 1.98261 = 6000 $$
$$ P = \frac{6000}{1.98261} \approx 3026.45 $$

Therefore, the size of each equal payment is approximately $3,026.45.

Question

To solve this problem, we need to determine the size of the equal payments that will settle the loan of $6,000, given that the payments are due in 2 months and 4 months, and the interest rate is 3.50% per annum. We will use the loan date as the focal date.

Step 1: Convert the annual interest rate to a monthly interest rate.
The annual interest rate is 3.50%. To find the monthly interest rate, we divide by 12:
$$ \text{Monthly interest rate} = \frac{3.50\%}{12} = \frac{0.035}{12} \approx 0.0029167 $$

Step 2: Calculate the present value of the payments.
Let $ P $ be the size of each equal payment. The present value of the first payment (due in 2 months) and the second payment (due in 4 months) must equal the loan amount of $6,000.

The present value of the first payment (due in 2 months) is:
$$ PV_1 = \frac{P}{(1 + 0.0029167)^2} $$

The present value of the second payment (due in 4 months) is:
$$ PV_2 = \frac{P}{(1 + 0.0029167)^4} $$

Step 3: Set up the equation for the total present value.
The total present value of the payments must equal the loan amount:
$$ PV_1 + PV_2 = 6000 $$
$$ \frac{P}{(1 + 0.0029167)^2} + \frac{P}{(1 + 0.0029167)^4} = 6000 $$

Step 4: Simplify the equation.
First, calculate the denominators:
$$ (1 + 0.0029167)^2 \approx 1.00584 $$
$$ (1 + 0.0029167)^4 \approx 1.01171 $$

Now, substitute these values into the equation:
$$ \frac{P}{1.00584} + \frac{P}{1.01171} = 6000 $$

Step 5: Combine the terms.
$$ \frac{P}{1.00584} + \frac{P}{1.01171} = 6000 $$
$$ P \left( \frac{1}{1.00584} + \frac{1}{1.01171} \right) = 6000 $$

Step 6: Calculate the sum of the fractions.
$$ \frac{1}{1.00584} \approx 0.99419 $$
$$ \frac{1}{1.01171} \approx 0.98842 $$
$$ 0.99419 + 0.98842 \approx 1.98261 $$

Step 7: Solve for $ P $.
$$ P \times 1.98261 = 6000 $$
$$ P = \frac{6000}{1.98261} \approx 3026.45 $$

Therefore, the size of each equal payment is approximately $3,026.45.

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