How many interest periods if you always pay double the interest A loan of $12,500$ is made at an effective interest rate of $8.5\%$. Payments are made at the end of each interest period. Each payment equals twice the interest due until the borrower pays off the outstanding debt with a final payment of at most $1800$. Find the number of payments $n$ and the amount of the final payment.

WHAT I'VE DONE SO FAR:

So I assumed the $n$th payment is $1800$. Then, in period $n-1$, the interest is $900$ and the principle is $900$. I dont know what to do after that. I tried to keep doubling the amount but then I got $n=4$ although the correct answer is $n=24$.
At time $n=0$ you have a loan $L=12,500$ to be repayed in $n$ periods with last  cumulative payment of $F=1800$. Let be, at time $n$, $P_n$ the payment, $C_n$ the principal repayed, $I_n$ the interest payed and $D_n$ the residual debt with $D_0=L$.

At each time $n$ you have the payment $$P_n=C_n+I_n=2I_n=2iD_{n-1}$$ and the remaining debt
$$
D_n=D_{n-1}-C_n=D_{n-1}-iD_{n-1}=iD_{n-1}
$$
thus we have
$$
\begin{align}
D_n&amp;=D_{0}(1-i)^n=L(1-i)^n\\
P_n&amp;=P_1(1-i)^{n-1}=2iL(1-i)^{n-1}
\end{align}
$$
At time $n$ we must have to pay at most $F=1800$, that is $P_n+D_n\le F$ and thus
$$
L\Big[2i(1-i)^{n-1}+(1-i)^n\Big]\le F\quad\Longrightarrow\quad (1-i)^n\le \frac{F}{L}\frac{1-i}{1+i}
$$
Finally we find 
$$
n\ge\frac{\log\left(\frac{F}{L}\frac{1-i}{1+i}\right)}{\log(1-i)}= 23.73 
$$
that is
$$
\boxed{n=24}
$$
So at year $n=24$ you will stop your loan and the final payment you will pay is the instalment $P_n$ and the lump sum $D_n$, that is $$P_{24}+D_{24}=275.45+1,482.57= 1,758.02&lt; 1,800$$

Just to check here is the amortization table:


n       P           C           I            D 
0                                        12,500.00 
1    2,125.00    1,062.50    1,062.50    11,437.50 
2    1,944.38      972.19      972.19    10,465.31 
3    1,779.10      889.55      889.55     9,575.76 
4    1,627.88      813.94      813.94     8,761.82 
5    1,489.51      744.75      744.75     8,017.07 
6    1,362.90      681.45      681.45     7,335.62 
7    1,247.05      623.53      623.53     6,712.09 
8    1,141.06      570.53      570.53     6,141.56 
9    1,044.07      522.03      522.03     5,619.53 
10     955.32      477.66      477.66     5,141.87 
11     874.12      437.06      437.06     4,704.81 
12     799.82      399.91      399.91     4,304.90 
13     731.83      365.92      365.92     3,938.98 
14     669.63      334.81      334.81     3,604.17 
15     612.71      306.35      306.35     3,297.82 
16     560.63      280.31      280.31     3,017.50 
17     512.98      256.49      256.49     2,761.01 
18     469.37      234.69      234.69     2,526.33 
19     429.48      214.74      214.74     2,311.59 
20     392.97      196.49      196.49     2,115.10 
21     359.57      179.78      179.78     1,935.32 
22     329.00      164.50      164.50     1,770.82 
23     301.04      150.52      150.52     1,620.30 
24     275.45      137.73      137.73     1,482.57
Okay, this is the amortization of a debt.

Payments are made at the end of every period, so at the time of your first payment you have a debt 12500(0.085)= 13562.5

Denote $Bn$ outstanding balance, $Pn$ principal payment and $Kn$ payment at period n and $i$ the interest.
$Bo$= 13567.5, interest for the initial period is $i Bo$ = (0.085) (13567.5)
implies $k1=2iBo$ since the paymets are twice the interest due amount.
and $B1= Bo- iBo$. One $Bo$ is to pay the debt and the other is to pay the interest.
Follow this we have: $B2= B1- iB1= Bo- 2iBo + i^3$ = $Bo * (1-2i + i^3)$
Now you see it? Binomial Theorem.

We want find n such that $Bn=Bo* (1-i)^n = 1800$ , ie , our Outstanding Balance must be 1800 to pay with a final payment
takin logarithm, n= 22.7 but 1800 AT MOST so take n=23 and the final payment is less than 1800. The answer is 23 payments and 1 final payment less that 1800.