An entrepreneur takes out a business loan for 60,000 with a nominal annual interest rate compounded monthly. The loan is scheduled to be paid off with level monthly payments, plus a final drop payment. All payments will be made at the end of the month. The principal portion of the payment is 1,400 for the first month and 1,414 for the second month.

Calculate the drop payment.

• 780
• 788
• 1183
• 1676
• 1692

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: E

Let j be the monthly rate and X be the level monthly payment. The principal repaid in the first payment is 1400 = X – 60,000j. The principal repaid in the second payment is 1414 = X –(60,000 – 1400)j. Substituting X = 1400 + 60,000j from the first equation gives 1414 = 1400 + 60,000j – 58,600j or 14 = 1400j and thus j = 0.01 and X = 2000. Let n be the number of payments. Then

[[math]]60000 = 2000 a_{\overline{n}|0.01}[[/math]]

and the calculator (or algebra) gives n = 35.8455. The equation for the drop payment, P, is

[[math]] 60,000=2000a_{\overline{35}|0.01}+P\nu^{36}=58,817.16+0.698925P [[/math]]

for P = 1962.

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.