The amortization amount in the first term is #\euro# #21500#

We have to write off #\euro \, 110000-30000=80000#. This is written off over #10# years. We will name the installment #a#. Since the installment decreases by #\euro \, 3000# each time, we must have:
\[a+(a-3000)+(a-6000)+\cdots+(a-27000)=80000\]
On the left hand side of the equation we have the sum of #n=10# terms of an arithmetic sequence with #v=-3000#. Moreover we have #t_1=a# and #t_{10}=a-27000#. According to the sum formula for an arithmetic sequence the left hand side is equal to \[\frac{1}{2} \cdot n \cdot (t_1+t_n)=\frac{1}{2} \cdot 10 \cdot(a+a-27000)=10\cdot a-135000\]
Hence, we have:
\[10\cdot a-135000=80000\]
This is a linear equation with unknown #a#. It can be solved as follows:
\[\begin{array}{rcl}
10\cdot a-135000&=&80000\\
&& \phantom{xxxxx}\color{blue}{\text{the original equation}}\\
10 \cdot a&=&215000\\
&& \phantom{xxxxx}\color{blue}{\text{added } 135000 \text{ to both sides}}\\
a &=& 21500\\
&& \phantom{xxxxx}\color{blue}{\text{both sides divided by }10}\\
\end{array}\]

We conclude that the amortization amount in the first term is #\euro \, 21500#.