Answer
$\$9088.67$
Work Step by Step
Let's note by $A_n$ the balance after $n$ months.
a) At the end of the first month she pays $200$ from a total of $10,000+0.005(10000)=1.005A_0$, so
$$A_1=1.005A_0-200.$$
At the end of the $n$th month she pays $200$ from a total of $A_{n-1}+0.005A_{n-1}=1.005A_{n-1}$, so
$$A_n=1.005A_{n-1}-200.$$
b) We determine $A_n$:
$$\begin{align*}
A_n&=1.005A_{n-1}-200\\
&=1.005(1.005A_{n-2}-200)-200\\
&=1.005^2A_{n-2}-1.005(200)-200\\
&=1.005^2(1.005A_{n-3}-200)-1.005(200)-200\\
&=1.005^3A_{n-3}-1.005^2(200)-1.005(200)-200\\
&=.......\\
&=1.005^nA_{0}-1.005^{n-1}(200)-1.005^{n-2}(200)-\dots -200\\
&=1.005^nA_{0}-200(1.005^{n-1}+1.005^{n-2}+\dots+1)\\
&=1.005^nA_{0}-200\cdot\dfrac{1.005^n-1}{1.005-1}\\
&=1.005^nA_{0}-40,000(1.005^n-1).
\end{align*}$$
We calculate $A_6$:
$$A_6=1.005^6(10,000)-40,000(1.005^6-1)\approx 9088.67.$$
