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 #20# is #\euro \, 20000#. 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# #40750#
We have to write off #\euro \, 360000-20000=340000#. This is written off over #20# years. We will name the installment #a#. Since the installment decreases by #\euro \, 2500# each time, we must have:
\[a+(a-2500)+(a-5000)+\cdots+(a-47500)=340000\]
On the left hand side of the equation we have the sum of #n=20# terms of an arithmetic sequence with #v=-2500#. Moreover we have #t_1=a# and #t_{20}=a-47500#. According to the sum formula for an arithmetic sequence the left hand side is equal to \[\frac{1}{2} \cdot n \cdot (t_1+t_n)=\frac{1}{2} \cdot 20 \cdot(a+a-47500)=20\cdot a-475000\]
Hence, we have:
\[20\cdot a-475000=340000\]
This is a linear equation with unknown #a#. It can be solved as follows:
\[\begin{array}{rcl}
20\cdot a-475000&=&340000\\
&& \phantom{xxxxx}\color{blue}{\text{the original equation}}\\
20 \cdot a&=&815000\\
&& \phantom{xxxxx}\color{blue}{\text{added } 475000 \text{ to both sides}}\\
a &=& 40750\\
&& \phantom{xxxxx}\color{blue}{\text{both sides divided by }20}\\
\end{array}\]
We conclude that the amortization amount in the first term is #\euro \, 40750#.
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