excans:D8cc90787d: Difference between revisions
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(Created page with "Find accumulated amount of groups of 4 payments at years <math>4,8,12,16, \ldots, 40</math>. Call them <math>K_1, K_2, \ldots, K_{10}</math>. Then find PV. <math>\left\{K_1, \ldots, K_{10}\right\}=\left\{s_{\overline{4} \mid .05}, 1.1025 s_{\overline{4} \mid .05}, \ldots, 1.1025^9 s_{\overline{4} \mid .05}\right\}</math>. Note that <math>1.05^2=1.1025</math>. Thus <math display="block"> \begin{aligned} & P V=K_1 1.05^{-4}+K_2 1.05^{-8}+\cdots=K_1 1.1025^{-2}+K_2 1.1025^...") |
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'''Solution: C''' | |||
Find accumulated amount of groups of 4 payments at years <math>4,8,12,16, \ldots, 40</math>. Call them <math>K_1, K_2, \ldots, K_{10}</math>. Then find PV. | Find accumulated amount of groups of 4 payments at years <math>4,8,12,16, \ldots, 40</math>. Call them <math>K_1, K_2, \ldots, K_{10}</math>. Then find PV. | ||
<math>\left\{K_1, \ldots, K_{10}\right\}=\left\{s_{\overline{4} \mid .05}, 1.1025 s_{\overline{4} \mid .05}, \ldots, 1.1025^9 s_{\overline{4} \mid .05}\right\}</math>. Note that <math>1.05^2=1.1025</math>. Thus | <math>\left\{K_1, \ldots, K_{10}\right\}=\left\{s_{\overline{4} \mid .05}, 1.1025 s_{\overline{4} \mid .05}, \ldots, 1.1025^9 s_{\overline{4} \mid .05}\right\}</math>. Note that <math>1.05^2=1.1025</math>. Thus | ||
Latest revision as of 23:40, 26 November 2023
Solution: C
Find accumulated amount of groups of 4 payments at years [math]4,8,12,16, \ldots, 40[/math]. Call them [math]K_1, K_2, \ldots, K_{10}[/math]. Then find PV. [math]\left\{K_1, \ldots, K_{10}\right\}=\left\{s_{\overline{4} \mid .05}, 1.1025 s_{\overline{4} \mid .05}, \ldots, 1.1025^9 s_{\overline{4} \mid .05}\right\}[/math]. Note that [math]1.05^2=1.1025[/math]. Thus
[[math]]
\begin{aligned}
& P V=K_1 1.05^{-4}+K_2 1.05^{-8}+\cdots=K_1 1.1025^{-2}+K_2 1.1025^{-4}+\ldots \\
& =s_{\overline{4} \mid .05}\left(1.1025^{-2}+1.1025^{-3}+\cdots+1.1025^{-11}\right)=s_{\overline{4} \mid .05}1.1025^{-1}\left(1.1025^{-1}+\cdots+1.1025^{-10}\right) \\
& =s_{\overline{4} \mid .05} 1.1025^{-1} a_{\overline{10} \mid .1025}=23.76580
\end{aligned}
[[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.