You could use Excel (see below) or you could solve the equation $(2)$ below numerically, e.g. using the secant method.
We have a so called uniform series of $n=60$ constant installments $m=400$.
Let $i$ be the nominal annual interest rate. The interest is compounded monthly, which means that the number of compounding periods per year is $12$. Consequently, the monthly installments $m$ are compounded at the interest rate per month $i/12$. The value of $m$ in the month $k$ is equivalent to the present value $m/(1+i/12)^{k}$. Summing in $k$, from $1$ to $n$, we get a sum that should be equal to $$P=26000-\frac{26000}{4}=19500.$$ This sum is the sum of a geometric progression of $n$ terms, with ratio $1+i/12$ and first term $m/(1+i/12)$. So
$$\begin{equation*}
P=\sum_{k=1}^{n}\frac{m}{\left( 1+\frac{i}{12}\right) ^{k}}=\frac{m}{1+\frac{
i}{12}}\frac{\left( \frac{1}{1+i/12}\right) ^{n}-1}{\frac{1}{1+i/12}-1}=m
\frac{\left( 1+\frac{i}{12}\right) ^{n}-1}{\frac{i}{12}\left( 1+\frac{i}{12}
\right) ^{n}}.\tag{1}
\end{equation*}$$
The ratio $P/m$ is called the series present-worth factor (uniform series)$^1$.
For $P=19500$, $m=400$ and $n=5\times 12=60$ we
have:
$$\begin{equation*}
19500=400
\frac{\left( 1+\frac{i}{12}\right) ^{60}-1}{\frac{i}{12}\left( 1+\frac{i}{12}
\right) ^{60}}.\tag{2}
\end{equation*}$$
I solved numerically $(2)$ for $i$ using SWP and got
$$
\begin{equation*}
i\approx 0.084923\approx 8.49\%.\tag{3}
\end{equation*}
$$
ADDED. Computation in Excel for the principal $P=19500$ and interest rate $i=0.084923$ computed above. I used a Portuguese version, that's why the decimal values show a comma instead of the decimal point.
- The Column $k$ is the month ($1\le k\le 60$).
- The 2nd. column is the amount $P_k$ still to be payed at the beginning of month $k$.
- The 3rd. column is the interest $P_ki/12$ due to month $k$.
- The 4th. column is the sum $P_k+P_ki/12$.
- The 5th column is the installment payed at the end of month $k$.
The amount $P_k$ satisfies $$P_{k+1}=P_k+P_ki/12-m.$$ We see that at the end of month $k=60$, $P_{60}+P_{60}i/12=400=m$. The last installment $m=400$ at the end of month $k=60$ balances entirely the remaining debt, which is also $400$. We could find $i$ by trial and error. Start with $i=0.01$ and let the spreadsheet compute the table values, until we have in the last row exactly $P_{60}+P_{60}i/12=400$.
--
$^1$ James Riggs, David Bedworwth and Sabah Randdhava, Engineering Economics,McGraw-Hill, 4th. ed., 1996, p.43.
Best Answer
If can't be calculated, because there is no data. We know, what was the inflation rate in 1994, what is it now, but we don't have data, what was between them. And it can't even be calculated out from the other data given in the question.