Accumulated Value of Investment So I have been given this questions
€500 is invested over a period of 4-years. In year 1 a nominal rate of interest of 6%
p.a. convertible quarterly applies. In year 2 a nominal rate of discount of 10% p.a.
convertible 4-monthly applies. In year 3 a force of interest of 0.75% per month applies.
In year 4 a compound rate of interest of 10% per 2-years applies. Calculate the
accumulated value of the investment at the end of the 4-year period and the
equivalent level annual force of interest over the 4-year period.
I started with year 1
$$i^{(4)} = 6\text{%} $$ $$p=4 $$
from this 
$$ i^{*} = \frac{i^{(p)}}{p} $$
$$ i^{*} = \frac{i^{(4)}}{4} = 0.015 $$
$$ AV_{1} = 500(1+i^{*})^{4} = 530.6817753 $$ 
We then move to year 2 
- In year 2 a nominal rate of discount of 10% p.a.
convertible 4-monthly applies.
$$ p = 3 $$
this is where my confusion arises, as it says discount rate.
I assume I must discount back some payments, however i am unsure of which payments that it is I am supposed to discount back
 A: Let $A(n)$ the accumulated value at the end of year $n$, starting from $A(0)=€ \,500$.


*

*$i_1^{(4)}=6\%$, so we have
$$
A(1)=A(0)\left(1+\frac{i_1^{(4)}}{4}\right)^4= 530.68
$$

*$d_2^{(3)}=10\%$, so we have
$$
A(2)=\frac{A(1)}{\left(1-\frac{d_2^{(3)}}{3}\right)^3}=587.49
$$

*$\delta_3=0.75\%$, so we have
$$
A(3)=A(2)\;\textrm{e}^{\delta_3 \times 12}=642.82
$$

*$i_4^{(1/2)}=10\%$, so we have
$$
A(4)=A(3)\left(1+\frac{i_1^{(1/2)}}{1/2}\right)^{1/2}=704.18
$$
So we have
$$
A(4)=A(0)\underbrace{\frac{\left(1+\frac{i_1^{(4)}}{4}\right)^4}{\left(1-\frac{d_2^{(3)}}{3}\right)^3}\;\textrm{e}^{\delta_3 \times 12}\;\left(1+\frac{i_1^{(1/2)}}{1/2}\right)^{1/2}}_{1.408351517}=704.18
$$
and we find the equivalent force of interest $\delta$ as $A(4)=A(0)\;\textrm{e}^{\delta \times 4}$, i.e.
$$
\delta=\frac14\times\log\left(\tfrac{\left(1+\frac{i_1^{(4)}}{4}\right)^4}{\left(1-\frac{d_2^{(3)}}{3}\right)^3}\;\textrm{e}^{\delta_3 \times 12}\;\left(1+\frac{i_1^{(1/2)}}{1/2}\right)^{1/2}\right)=\frac14\times\log(1.408351517)\approx 8.56\%
$$
