Arithmetic-Increasing Perpetuity with Multi-Year Spacing
This Exam FM sample reference tests Arithmetic-Increasing Perpetuity with Multi-Year Spacing. Treat each three-year block as one period. Its effective rate is 19.1016%, and the increasing-perpetuity factor is 1/J plus 1/J squared. Solving the 655.56 price gives X = 20.083, choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AChoice A discounts the nominal sum as though only one payment occurred every three years.
BChoice B values a level perpetuity of X and omits the arithmetic increase.
DChoice D uses 18% as a simple three-year rate rather than the effective 19.1016% rate.
EChoice E places the first payment at year 1 instead of year 3.
Original practice · fully worked
Original variant: infer annual yield from a spaced arithmetic perpetuity
A perpetuity pays 10 at year 3, 20 at year 6, 30 at year 9, and continues increasing by 10 every three years. Its present value is 300. Determine the annual effective yield.
A 5.00%
B 5.74%
C 6.27%
D 6.75%
E 7.20%
Variant answer in brief
The three-year rate solving 300 = 10(1/J + 1/J squared) is 20%. Converting that rate to an annual effective yield gives 6.2659%, choice C.
Setup
Setup
Let J be the effective yield over one three-year payment interval.
300=10(J1+J21)
Model
Model
Solve the positive quadratic root.
30J2−J−1=0
J=0.20
Compute
Compute
Convert the three-year rate to an annual effective yield.
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