Sequences and series: Financial applications of sequences and series
Other applications of sequences and series
Up to now we've already seen a couple of applications of arithmetic and geometric sequences. Now we will look at some more examples of arithmetic sequences and series from the amortization of a new machine.
The residual value of that machine after year #5# is #\euro \, 10000#. Possible interest expenses are not considered.
Calculate the amount of the amortization in the first term.
The amortization amount in the first term is #\euro# #77000#
We have to write off #\euro \, 370000-10000=360000#. This is written off over #5# years. We will name the installment #a#. Since the installment decreases by #\euro \, 2500# each time, we must have:
\[a+(a-2500)+(a-5000)+\cdots+(a-10000)=360000\]
On the left hand side of the equation we have the sum of #n=5# terms of an arithmetic sequence with #v=-2500#. Moreover we have #t_1=a# and #t_{5}=a-10000#. According to the sum formula for an arithmetic sequence the left hand side is equal to \[\frac{1}{2} \cdot n \cdot (t_1+t_n)=\frac{1}{2} \cdot 5 \cdot(a+a-10000)=5\cdot a-25000\]
Hence, we have:
\[5\cdot a-25000=360000\]
This is a linear equation with unknown #a#. It can be solved as follows:
\[\begin{array}{rcl}
5\cdot a-25000&=&360000\\
&& \phantom{xxxxx}\color{blue}{\text{the original equation}}\\
5 \cdot a&=&385000\\
&& \phantom{xxxxx}\color{blue}{\text{added } 25000 \text{ to both sides}}\\
a &=& 77000\\
&& \phantom{xxxxx}\color{blue}{\text{both sides divided by }5}\\
\end{array}\]
We conclude that the amortization amount in the first term is #\euro \, 77000#.
