Revision as of 00:08, 27 November 2023 by Admin (Created page with "'''Solution: A''' <math>108=K\left(1-v^{14}\right)</math> <math display="block"> 135=K\left(1-v^{28}\right)=K\left(1-v^{14}\right)\left(1+v^{14}\right)=108\left(1+v^{14}\right) \text { so } 1+v^{14}=135 / 108=1.25 \text { so } v^{14}=.25 \text {. } </math> Thus <math>K=108 /\left(1-v^{14}\right)=108 /(1-.25)=144</math>. Interest in 29th installment is <math>K\left(1-v^7\right)=144(1-\sqrt{.25})=72</math>. '''References''' {{cite web |url=https://web2.uwindsor.ca/ma...")
Exercise
ABy Admin
Nov 27'23
Answer
Solution: A
[math]108=K\left(1-v^{14}\right)[/math]
[[math]]
135=K\left(1-v^{28}\right)=K\left(1-v^{14}\right)\left(1+v^{14}\right)=108\left(1+v^{14}\right) \text { so } 1+v^{14}=135 / 108=1.25 \text { so } v^{14}=.25 \text {. }
[[/math]]
Thus [math]K=108 /\left(1-v^{14}\right)=108 /(1-.25)=144[/math]. Interest in 29th installment is [math]K\left(1-v^7\right)=144(1-\sqrt{.25})=72[/math].
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.