The amortization amount in the first term is #\euro# #45750#

We have to write off #\euro \, 420000-30000=390000#. This is written off over #10# years. We will name the installment #a#. Since the installment decreases by #\euro \, 1500# each time, we must have:
\[a+(a-1500)+(a-3000)+\cdots+(a-13500)=390000\]
On the left hand side of the equation we have the sum of #n=10# terms of an arithmetic sequence with #v=-1500#. Moreover we have #t_1=a# and #t_{10}=a-13500#. According to the sum formula for an arithmetic sequence the left hand side is equal to \[\frac{1}{2} \cdot n \cdot (t_1+t_n)=\frac{1}{2} \cdot 10 \cdot(a+a-13500)=10\cdot a-67500\]
Hence, we have:
\[10\cdot a-67500=390000\]
This is a linear equation with unknown #a#. It can be solved as follows:
\[\begin{array}{rcl}
10\cdot a-67500&=&390000\\
&& \phantom{xxxxx}\color{blue}{\text{the original equation}}\\
10 \cdot a&=&457500\\
&& \phantom{xxxxx}\color{blue}{\text{added } 67500 \text{ to both sides}}\\
a &=& 45750\\
&& \phantom{xxxxx}\color{blue}{\text{both sides divided by }10}\\
\end{array}\]

We conclude that the amortization amount in the first term is #\euro \, 45750#.