Ans: A
$\begin{array}{rcl}
P(1+\dfrac{6\%}{12})^{12} + P(1+\dfrac{6\%}{12})^{11} + \cdots + P(1+\dfrac{6\%}{12}) & = & 10~000 \\
P(1.005)\times \dfrac{1.005^{12}-1}{1.005-1} & = & 10~000 \\
P & \approx & 806.63
\end{array}$
$\begin{array}{rcl}
P(1+\dfrac{6\%}{12})^{12} + P(1+\dfrac{6\%}{12})^{11} + \cdots + P(1+\dfrac{6\%}{12}) & = & 10~000 \\
P(1.005)\times \dfrac{1.005^{12}-1}{1.005-1} & = & 10~000 \\
P & \approx & 806.63
\end{array}$
