Revision as of 08:45, 18 November 2023 by Admin (Created page with "<math display = "block"> \begin{align*} 167.50 &=10a_{\overline{5}|9.2\%}+10(1.092)^{-5}\sum_{t=1}^{c}\left[\frac{(1+k)}{1\cdot092}\right]^{t} \\ 167.50 &=38.6955+6.44001{\frac{(1+k)/1.092}{1-(1+k)/1.092}} \\ (167.50-38.6955)[1-(1+k)/1.092] &= 6.4400](1+k)/1.092 \\ 128.8045 &= 135.24451(1+k)/1.092 \\ 1+k &= 0.0400 \\ k &= 0.0400 \Rightarrow K= 4.0\% \end{align*} </math> {{soacopyright | 2023 }}")
Exercise
ABy Admin
Nov 18'23
Answer
[[math]]
\begin{align*}
167.50 &=10a_{\overline{5}|9.2\%}+10(1.092)^{-5}\sum_{t=1}^{c}\left[\frac{(1+k)}{1\cdot092}\right]^{t} \\
167.50 &=38.6955+6.44001{\frac{(1+k)/1.092}{1-(1+k)/1.092}} \\
(167.50-38.6955)[1-(1+k)/1.092] &= 6.4400](1+k)/1.092 \\
128.8045 &= 135.24451(1+k)/1.092 \\
1+k &= 0.0400 \\
k &= 0.0400 \Rightarrow K= 4.0\%
\end{align*}
[[/math]]
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