during the first 4 years, interest is credited using a simple I'm having trouble with the following problem from actuarial exam FM: 
During the first 4 years, interest is credited using a simple interest rate of $5\%$ a year. After 4 years, interest is credited at a force of interest: 
$$\delta_t = \frac{0.2}{1+0.2t}, t \geq 4$$ 
The following are numerically equal: 
(i) the current value at time $t = 4$ of payments of 1000 at time $t =2$ and 400 at time $t = 7$; and (ii) the present value at time $t = 0$ of a payment of $X$ at time $t = 10$. 
I have two questions about the solution


*

*The solution says the current value of (i) $= 1000[1 + 2(.05)] + 400\frac{a(4)}{a(7)}$. I was wondering why the first term isn't $1000\frac{a(4)}{a(2)}$. 

*I thought that the value of (ii) would be $X\cdot \frac{1}{1+.05(4)} \cdot \frac{1.8}{1+.2(6)}$, where the last fraction is the inverted $a(t)$ you get from the force of interest. But the solutions say something different. I'm wondering why my representation is not correct.


Thank you very much for your help in advance!
 A: For the point (i)


*

*The current value $V'$ at time $t=4$ of payments of $1000$ at time $t=2$ is the future value of $1000$ at the simple interest rate $i=5\%$ for $2$ years using the formula $a(n)=a(0)(1+in)$
$$
V'=1000\,(1+2\times 5\%)=1000\times 1.1
$$
and $1.1=(1+2\times 5\%)=\frac{a(4)}{a(2)}$.

*The current value $V''$ at time $t=4$ of payments of $400$ at time $t=7$ is the present value of $400$ at the force of interest rate $\delta_t$ for $3$ years using the formula $a(t)=a(t_0)\mathrm{e}^{\int_{t_0}^t\delta_\tau\mathrm d \tau}$. Observing that $\mathrm{e}^{\int_{t_0}^t\delta_\tau\mathrm d \tau}=\mathrm{e}^{\int_{t_0}^t\frac{0.2}{1+0.2\tau}\mathrm d \tau}=\mathrm{e}^{\left(\log(\tau+5)\big|_{t_0}^t\right)}=\frac{t+5}{t_0+5}$, we have $\frac{a(t)}{a(t_0)}=\frac{t+5}{t_0+5}$
$$
V''=400\times\frac{a(4)}{a(7)}=400\times \frac{4+5}{7+5}=400\times \frac{9}{12}
$$

*The current value at $t=4$ then is 
$$
V=V'+V''=1400
$$
For the point (ii)
The present value $W$ at time $t=0$ of a payment of $X$ at time $t=10$ is the discounted value $W'=X\cdot\frac{a(4)}{a(10)}$ at the interest force $\delta_t$ at time $t=4$, which is then discounted at the simple interest $i$
$$
W=\frac{W'}{1+4i}=X\cdot\frac{1}{1+4i}\cdot\frac{a(4)}{a(10)}
$$ 
that is
$$
W=X\cdot \frac{1}{1+4\times 0.05}\cdot \frac{4+5}{10+5}= X\cdot \frac{1}{1.2}\cdot \frac{9}{15}=X\cdot \frac{0.6}{1.2}=\frac{X}{2}
$$
Find $X$
We know that $V=W$, so we have 
$$
1400=\frac{X}{2}\quad\Longrightarrow\quad \boxed{X=2800}
$$
