Determining outstanding balance using the prospective method 
A loan is being repaid with 10 payments of 2,000 each followed by
  10 payments of 1,000 each at the end of each half-year. Assume
  that the nominal rate of interest convertible semiannually equals
  $i^{(2)} = 10%$.
  Find the outstanding loan balance immediately after the fifth
  payment is made by 
(1) the prospective method 
(2) the retrospective method

For the prospective method, the solution shows
$B_5^p = 1000(a_{15|0.05}+a_{5|0.05})$ 
I know that the formula of the prospective method is $B_5^p = a_{n-t}$, so where does the $ + (a_{5|0.05})$ come from? Why isn't it just $B_5^p = 1000(a_{15|0.05})$
 A: Immediately after the fifth payment (at $t=5$), you made $5$ payments of $2000$ and you have to pay $5$ payments of $2000$ beginning at $t=6$ and $10$ payments of $1000$ beginning at $t=11$ (i.e. deferred of 10), that is 
$$
B_5^p=2000\,a_{\overline{5}|j}+1000\,_{10|}a_{\overline{5}|j}\tag 1
$$
and observing that the deferred annuity satisfies $_{m|}a_{\overline{n}|j}=v^m a_{\overline{n}|j}=a_{\overline{n+m}|j}-a_{\overline{m}|j}$, the $(1)$ becomes
$$
B_5^p=2000\,a_{\overline{5}|j}+1000\,(a_{\overline{15}|j}-a_{\overline{5}|j})=1000\,a_{\overline{15}|j}+1000\,a_{\overline{5}|j}\tag 2
$$
As second way of solution, observe that the stream of payments is equivalent to a sum of 15 payments of 1000  beginning at $t=6$ and 5 payments of 1000 beginning at $t=6$.
Then at $t=5$ you have immediately
$$
B_5^p=1000\,a_{\overline{15}|j}+1000\,a_{\overline{5}|j}=1000(a_{\overline{15}|j}+a_{\overline{5}|j})\tag 3
$$
For the retrospective method you can find easily that the loan $L$ is $$L=2000 a_{\overline{10}|j}+1000\, _{10|}a_{\overline{10}|j}=1000 (a_{\overline{20}|j}+a_{\overline{10}|j})\tag 4$$
and then
$$
B_5^r=L(1+j)^5-2000\,s_{\overline{5}|j}=1000 (a_{\overline{20}|j}+a_{\overline{10}|j})(1+j)^5-2000\,s_{\overline{5}|j}\tag 5
$$
Observing that $$(1+j)^ma_{\overline{n}|j}=v^{n-m}s_{\overline{n}|j}=s_{\overline{m}|j}+a_{\overline{n-m}|j}$$ the $(5)$ becomes
$$
\begin{align}
B_5^r&=1000 (s_{\overline{5}|j}+a_{\overline{15}|j}+s_{\overline{5}|j}+a_{\overline{5}|j})-2000\,s_{\overline{5}|j}\\
&=2000\,s_{\overline{5}|j}+1000 (a_{\overline{15}|j}+a_{\overline{5}|j})-2000\,s_{\overline{5}|j}\\
&=1000 (a_{\overline{15}|j}+a_{\overline{5}|j})\tag 6
\end{align}
$$
that is equal to $(3)$.
