The cash flow is
$$
\begin{matrix}
\text{Payments } & 800 & 750 &700 & 650 & 600 & 550 & 500 & 450 & 400 & 350\\
\hline{}
\text{Time } & 1 & 2 & 3 & 4& 5 & 6 & 7 & 8 & 9 & 10
\end{matrix}\tag 1$$
and this is obvious that the cash flow decreases by $50$.
The cash flow $(1)$ is equivalent to the following cash flow
$$\begin{matrix}
\text{Payments } & 500 & 450 & 400 & 350 & 300 & 250 & 200 & 150 & 100 & 50\\
\hline{}
\text{Payments } & 300 & 300 & 300 & 300 & 300 & 300 & 300 & 300 & 300 & 300\\
\hline{}
\text{Time } & 1 & 2 & 3 & 4& 5 & 6 & 7 & 8 & 9 & 10
\end{matrix}\tag 2
$$
or to this cash flow
$$
\begin{matrix}
\text{Payments } & 450 & 400 & 350 & 300 & 250 & 200 & 150 & 100 & 50 & 0\\
\hline{}
\text{Payments } & 350 & 350 & 350 & 350 & 350 & 350 & 350 & 350 & 350 & 350\\
\hline{}
\text{Time } & 1 & 2 & 3 & 4& 5 & 6 & 7 & 8 & 9 & 10
\end{matrix}\tag 3
$$
Let be $i=\frac{i^{(2)}}{2}=8\%,\,n=10,\,Q=50,\, P=300,\,P'=P+Q=350$.
So for cashflow $(2)$ we have
$$
PV=P\,a_{\overline{n}|i}+Q(Da)_{\overline{n}|i}=300\,a_{\overline{10}|0.08}+50(Da)_{\overline{10}|0.08}\tag 4
$$
and for cashflow $(3)$ we have
$$
PV=P'\,a_{\overline{n}|i}+Q(Da)_{\overline{n-1}|i}=350\,a_{\overline{10}|0.08}+50(Da)_{\overline{9}|0.08}\tag 5
$$
It's obvious that $(5)=(4)$
\begin{align}
P'\,a_{\overline{n}|i}+Q(Da)_{\overline{n-1}|i}&=
(P+Q)\,a_{\overline{n}|i}+Q(Da)_{\overline{n-1}|i}\\
&=P\,a_{\overline{n}|i}+Q\,a_{\overline{n}|i}+Q(Da)_{\overline{n-1}|i}\\
&=P\,a_{\overline{n}|i}+Q\left[\,a_{\overline{n}|i}+(Da)_{\overline{n-1}|i}\right]\\
&=P\,a_{\overline{n}|i}+Q\left[\,a_{\overline{n}|i}+\frac{n-1-a_{\overline{n-1}|i}}{i}\right]\\
&=P\,a_{\overline{n}|i}+Q\left[\frac{i\,a_{\overline{n}|i}+n-(1+a_{\overline{n-1}|i})}{i}\right]\\
&=P\,a_{\overline{n}|i}+Q\left[\frac{n-a_{\overline{n}|i}}{i}\right]\\
&=P\,a_{\overline{n}|i}+Q(Da)_{\overline{n}|i}
\end{align}
using $1+a_{\overline{n-1}|i}=\ddot a_{\overline{n}|i}=(1+i)a_{\overline{n}|i}$. So you have also the result $(Da)_{\overline{n}|i}=a_{\overline{n}|i}+(Da)_{\overline{n-1}|i}$.
Other choises are possible. For example we can choose the following decomposition
$$\begin{matrix}
\text{Payments } & 0 & -50 & -100 & -150 & -200 & -250 & -300 & -350 & -400 & -450\\
\hline{}
\text{Payments } & 800 & 800 & 800 & 800 & 800 & 800 & 800 & 800 & 800 & 800\\
\hline{}
\text{Time } & 1 & 2 & 3 & 4& 5 & 6 & 7 & 8 & 9 & 10
\end{matrix}\tag 6
$$
So, putting $F=800$, the present value will be
$$
PV=F\,a_{\overline{n}|i}-Q\,v\,(Ia)_{\overline{n-1}|i}\tag 7
$$
We can show that $(7)=(2)$, observing that $F=P+nQ$,
\begin{align}
F\,a_{\overline{n}|i}-Q\,v\,(Ia)_{\overline{n-1}|i}&=
(P+nQ)\,a_{\overline{n}|i}-Q\,v\,(Ia)_{\overline{n-1}|i}\\
&=P\,a_{\overline{n}|i}+Q\left[n\,a_{\overline{n}|i}-v(Ia)_{\overline{n-1}|i}\right]\\
&=P\,a_{\overline{n}|i}+Q\left[n\,a_{\overline{n}|i}-v\frac{(1+i)a_{\overline{n-1}|i}-(n-1)v^{n-1}}{i}\right]\\
&=P\,a_{\overline{n}|i}+Q\left[n\,a_{\overline{n}|i}-\frac{a_{\overline{n-1}|i}-(n-1)v^{n}}{i}\right]\\
&=P\,a_{\overline{n}|i}+Q\left[n\,a_{\overline{n}|i}-\frac{(1+i)a_{\overline{n}|i}-1-(n-1)v^{n}}{i}\right]\\
&=P\,a_{\overline{n}|i}+Q\left[\frac{ni\,a_{\overline{n}|i}-a_{\overline{n}|i}-ia_{\overline{n}|i}+1+(n-1)v^{n}}{i}\right]\\
&=P\,a_{\overline{n}|i}+Q\left[\frac{(n-1)i\,a_{\overline{n}|i}-a_{\overline{n}|i}+1+(n-1)v^{n}}{i}\right]\\
&=P\,a_{\overline{n}|i}+Q\left[\frac{(n-1)(1-v^n)-a_{\overline{n}|i}+1+(n-1)v^{n}}{i}\right]\\
&=P\,a_{\overline{n}|i}+Q\left[\frac{n-a_{\overline{n}|i}}{i}\right]\\
&=P\,a_{\overline{n}|i}+Q\,(Da)_{\overline{n}|i}\\
\end{align}
using $1+a_{\overline{n-1}|i}=\ddot a_{\overline{n}|i}=(1+i)a_{\overline{n}|i}$.
