The amortization amount in the first term is #\euro# #47000#

We have to write off #\euro \, 400000-10000=390000#. This is written off over #15# years. We will name the installment #a#. Since the installment decreases by #\euro \, 3000# each time, we must have:
\[a+(a-3000)+(a-6000)+\cdots+(a-42000)=390000\]
On the left hand side of the equation we have the sum of #n=15# terms of an arithmetic sequence with #v=-3000#. Moreover we have #t_1=a# and #t_{15}=a-42000#. According to the sum formula for an arithmetic sequence the left hand side is equal to \[\frac{1}{2} \cdot n \cdot (t_1+t_n)=\frac{1}{2} \cdot 15 \cdot(a+a-42000)=15\cdot a-315000\]
Hence, we have:
\[15\cdot a-315000=390000\]
This is a linear equation with unknown #a#. It can be solved as follows:
\[\begin{array}{rcl}
15\cdot a-315000&=&390000\\
&& \phantom{xxxxx}\color{blue}{\text{the original equation}}\\
15 \cdot a&=&705000\\
&& \phantom{xxxxx}\color{blue}{\text{added } 315000 \text{ to both sides}}\\
a &=& 47000\\
&& \phantom{xxxxx}\color{blue}{\text{both sides divided by }15}\\
\end{array}\]

We conclude that the amortization amount in the first term is #\euro \, 47000#.