I have this question here that I'm having trouble understanding
An annuity immediate has semiannual payments of 800,750, 700,..., 350 at $i^{(2)}$ $= .16$ if $a_{\overline10|.08} = A$, find the present value of the annuity in terms of A
The solution is:
PV = $300a_{\overline10|.08}$ $+ 50(Da)_{\overline10|0.08}$
What I don't understand is why our annuity is 300 per payment, and why the decreasing annuity starts at 50?
It would make sense to me if the annuity was starting at 800 per payment, and decreasing by 50.
So something like PV = $800a_{\overline10|.08}$ $+ 50(Da)_{\overline10|0.08}$
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 $1+a_{\overline{n-1}|i}=\ddot a_{\overline{n}|i}=(1+i)a_{\overline{n}|i}$.