The amortization amount in the first term is #\euro# #12500#

We have to write off #\euro \, 100000-40000=60000#. This is written off over #20# years. We will name the installment #a#. Since the installment decreases by #\euro \, 1000# each time, we must have:
\[a+(a-1000)+(a-2000)+\cdots+(a-19000)=60000\]
On the left hand side of the equation we have the sum of #n=20# terms of an arithmetic sequence with #v=-1000#. Moreover we have #t_1=a# and #t_{20}=a-19000#. According to the sum formula for an arithmetic sequence the left hand side is equal to \[\frac{1}{2} \cdot n \cdot (t_1+t_n)=\frac{1}{2} \cdot 20 \cdot(a+a-19000)=20\cdot a-190000\]
Hence, we have:
\[20\cdot a-190000=60000\]
This is a linear equation with unknown #a#. It can be solved as follows:
\[\begin{array}{rcl}
20\cdot a-190000&=&60000\\
&& \phantom{xxxxx}\color{blue}{\text{the original equation}}\\
20 \cdot a&=&250000\\
&& \phantom{xxxxx}\color{blue}{\text{added } 190000 \text{ to both sides}}\\
a &=& 12500\\
&& \phantom{xxxxx}\color{blue}{\text{both sides divided by }20}\\
\end{array}\]

We conclude that the amortization amount in the first term is #\euro \, 12500#.