I'm going to formalize this a bit, but I will bring this back to earth, I promise. Assume we begin with a principal $P_0$, interest rate $i$ per compounding period (which is a month), and payment per compounding period $m$ (monthly payment). Also assume that there is another additional payment $q$ made every $\ell$ compounding periods (e.g., the additional annual payment, where $\ell=12$).
The way to express the amount owed after the first extra payment is made is to determine the present value $P_1$ of what is owed and what has been paid:
$$P_1 = \underbrace{\left [ \cdots \left [ \left [ P_0 (1+i) - m\right ] (1+i) - m \right ] (1+i) - m \cdots \right ] (1+i)}_{k \, \text{compunding periods}} - m - q$$
Collecting all the terms, we have
$$P_1 = P_0 (1+i)^k - m \sum_{j=0}^{k-1} (1+i)^j - q$$
Summing the geometric series, we finally get
$$P_1 = P_0 (1+i)^k - m \frac{(1+i)^k-1}{i} - q$$
To get the principal after $\ell$ such periods, we note that we may consider the beginning of the second period as having a principal equal to $P_1$; the beginning f the third period starts out having principal $P_2$, and so on. So we get a recurrence relation
$$P_{\ell} = P_{\ell-1} (1+i)^k - m \frac{(1+i)^k-1}{i} - q$$
This is a simple, inhomogeoneus, first-order recurrence in $P_{\ell}$ of the form
$$P_{\ell} - a P_{\ell-1} = b$$
The general solution to this equation is
$$P_{\ell} = C a^{\ell} - \frac{b}{a-1}$$
where
$$P_0 = C- \frac{b}{a-1}$$
Using $a=(1+i)^k$ and $b = - m \frac{(1+i)^k-1}{i} - q$, I get for the principal after $\ell$ periods:
$$P_{\ell} = P_0 (1+i)^{k \ell} - \frac{(1+i)^{k \ell}-1}{(1+i)^k-1}\frac{m [(1+i)^k-1]+q i}{i} $$
Now plug in the values $m=700$, $k=12$, $\ell=10$, $i=0.12/12=0.01$, $q=1000$; I get about
$$P_{10} \approx 3.3003869 P_0 - 179,165.88$$
($P_0$ was never specified.) If $P_{10} = 0$, then $P_0 \approx 77,884.88$.
