Prepaying interest with set installment amount where extra payment is applied to principal

Question

I believe this is a problem of computing the limit of a recursive sequence.

Let's say I have a mortgage of \$332950 and a 6% annual interest rate. The mortgage contract requires that interest for each year is prepaid for each year on Dec 31 in the amount of \$20,000. (Interest for 2016 is paid on Dec 31 2015) Any amount paid above the required interest is applied to the principal. Now, 6% of \$332950 is \$19977, NOT \$20000, so by default the contract requires paying more than just interest, and so some of that money will be applied to the principal.

The problem is that if \$23 of the $20000 installment is applied to principal, then the the interest due is thereby decreased. If the interest due is decreased, then even more is applied to principal. I ran a 100 iterations of this cycle using a python script, and after 10 iterations, the amount applied to interest reaches a limit of \$24.4680851064. How can I compute this limit without using brute force?

If I understand the notation correctly, this is how to define the sequence:

Suppose $a_0 = 0$, and $$ a_n = 20000 - ( ( 332950 - a_{n-1}) \cdot 0.06)$$ if $n \ge 1$.

How do I compute the limit as $n$ approaches $\infty$?

iteration   a
0           0
1           23
2           24.38
3           24.4628
4           24.467768
5           24.46806608
6           24.46808396
7           24.46808504
8           24.4680851
9           24.46808511
10          24.46808511
...         ...
100,000,000 24.46808511

Answer 1

Thanks to @SalahFatima for the comments. If I replace $a_n$ and $a_{n-1}$ with $L$, then I have the following solution:

$$ a_n = 20000 - ( ( 332950 - a_{n-1}) \cdot 0.06)$$ $$ L = 20000 - ( ( 332950 - L) \cdot 0.06)$$ $$ L = 20000 - 19977 + 0.06L$$ $$ L = 23 + 0.06L$$ $$ 0.94L = 23$$ $$ L = \frac{23}{0.94}$$ $$ L = 24.4680851 $$