Mary purchases an increasing annuity-immediate for 50,000 that makes twenty annual payments as follows:

1. P, 2P, . . . , 10P in years 1 through 10, and
2. 10P (1.05), 10P (1.05)², . . . , 10P (1.05)¹⁰ in years 11 through 20.

The annual effective interest rate is 7% for the first 10 years and 5% thereafter.

Calculate P

• 564,
• 574,
• 584,
• 594,
• 604

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

Solution: C

Let [math]v=1 /(1+.07)[/math] and let [math]w=1 /(1+.05)[/math].

[[math]] \begin{aligned} 50000 & =\left(P v+2 P v^2+\cdots+10 P v^{10}\right)+v^{10}\left(10 P(1.05) w+\cdots+10 P(1.05)^{10} w^{10}\right) \\ & =P v\left(1+2 v+\cdots+10 v^9\right)+v^{10} P(10+10+\cdots+10) \end{aligned} [[/math]]

[[math]] \begin{aligned} S & =1+2 v+3 v^2+\cdots+10 v^9 \\ v S & =1 v+2 v^2+\cdots+9 v^9+10 v^{10} \\ (1-v) S & =1+v+\cdots+v^9-10 v^{10} \end{aligned} [[/math]]

So

[[math]]S=\frac{1-v^{10}}{(1-v)^2}-\frac{10 v^{10}}{1-v}=\frac{1-v^{10}-10 v^{10}+10 v^{11}}{(1-v)^2}=\frac{1-5.59184+4.75093}{.00427985}=37.170874[[/math]]

so

[[math]]50000=P(1.07)^{-1}(37.170874)+(1.07)^{-10} P(100)[[/math]]

so [math]P=\frac{50000}{34.739135+50.8349292}=584.29 . [/math]

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.