Financial Mathematics, interest problem Determine the amount of interest earned from time $t=2$ to $t=4$ if $240$ is invested at $t=1$ and an additional $300$ is invested at $t=3$.
Given, $a(1)=1.2$, $a(2)=1.5$, $a(3)=2.0$, $a(4)=3.0$
I tried finding $A(0)$ using the equation $A(1)=A(0).a(1)$. Hence $A(0)=200$
Hence I found $A(2)=300$, $A(3)=400$, $A(4)=600$
$I_{[2,3]}=A(0)[\text{new } A(3)-A(2)]=80000$
$I_{[3,4]}=A(0)[A(4)-\text{new }A(3)]$
From here I am stuck.I am messed up with the different times.

Answer:
$$a_{(2-4)} = \frac{a(4)}{a(2)}$$
$$A(2) = A(1).\frac{a(2)}{a(1)}$$
$$A(2) =240\times\frac{1.5}{1.2} = 300$$
$$A(4) = 300\times\frac{3}{1.5} = 600$$
$$A'(4) = A'(3)\frac{a(4)}{a(3)}$$
$$A'(4) = 300\times\frac{3}{2}$$
$$A'(4) = 450$$
$$I_1 = A(4)-A(2) = 600-300 = 300$$
$$I_2 = A'(4) - A'(3) = 450-300 = 150$$
$$I = I_1 +I_2$$
The required Interest:
$$I = 300+150 = 450$$
Regards
Satish
