Question regarding interest method during a year. $i-\delta$ 
B and P deposits the amount $100$ in separate bank accounts.  B's account have a nominal rate convertible semiannually.  P's account has a force of interest $\delta$.  After $7.25$ years, each account has $200$.  Calculate $i-\delta$.

The following is what I tried.
Under P's account,
$$100\exp \left( \int_0^{7.25} \delta \quad dt \right) = 200$$
so we can solve for
$$\delta = \frac{\ln 2 }{7.25}$$
Which I understand.  What I am having trouble with is B's account.
I can see that for the first $7$ years under the assumption that the nominal annual rate is $i$, the initial value will be multiplied by $(1+i)^7$, but what happens for the rest $0.25$ of the year?
I've been hearing that during the year between the conversion period, usually people use simple interest to calculate the amount.  However, if we did that then $i$ will be calculated as
$$100(1+i)^7(1+0.25i)=200$$
which is beyond my level because I cannot use a graphing calculator to solve this problem.
Even assuming that since the conversion occurs $3$ months later and I am able to find that 
$$100(1+i)^7=200$$
then
$$i = \sqrt[7]{2}-1$$
my $i- \delta$ does not equal $0.23\%$, which is supposedly the correct answer.
I tried using $i^{(2)}$ where
$$100(1+ \frac{(i^{(2)}}{2})^2=200$$
but $i^{(2)}-\delta$ was not the answer either.
Cal I have some help?

I would calculate $C_{7.5}=100\cdot \left(1+\frac{i}{2}\right)^{14.5}=200$
$\text{\$}100$ are compounded with the interest rate of $\frac{i}{2}$ in the first six months. Then this amount of money is compounded for another 6 months with the interest rate of $\frac{i}{2}$. It goes on till the 15th half year.
The general formula is $C_n=C_0\cdot\left(1+\frac{i}{m}\right)^{m \cdot n}$
