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 $B_5^p = 1000(a_{15|0.05})$
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 $(3)$.