Revision as of 21:46, 18 November 2023 by Admin (Created page with "'''Solution: C''' The interest earned is a decreasing annuity of 6, 5.4, etc. Combined with the annual deposits of 100, the accumulated value in fund Y is <math display = "block"> \begin{align*} &= 6(D s)_{\overline{{{10}}}|0.09}+100s_{\overline{{{10}}}|0.09}\\ &=6\left(\frac{10\left(1.09\right)^{10}-s_{\overline{{{10}}}|0.09}}{0.09}\right)+100\bigl(15.19293\bigr) \\ &= 565.38 + 1519.29 \\ &= 2084.67. \end{align*} </math> {{soacopyright | 2023 }}")
Exercise
ABy Admin
Nov 18'23
Answer
Solution: C
The interest earned is a decreasing annuity of 6, 5.4, etc. Combined with the annual deposits of 100, the accumulated value in fund Y is
[[math]]
\begin{align*}
&= 6(D s)_{\overline{{{10}}}|0.09}+100s_{\overline{{{10}}}|0.09}\\
&=6\left(\frac{10\left(1.09\right)^{10}-s_{\overline{{{10}}}|0.09}}{0.09}\right)+100\bigl(15.19293\bigr) \\
&= 565.38 + 1519.29 \\
&= 2084.67.
\end{align*}
[[/math]]