Question

How do you write a rule for the $n^{th}$ term \[7,5,3,1,-1\] ?

Answer 1

Hint: In order to find a solution to this problem, we will have to analyze that our problem is a descending arithmetic sequence, that is, two consecutive terms differ by a common difference. In our case, two consecutive terms always differ by $2$, which means that if we know the ${{n}^{th}}$ terms, we will get the \[n+{{1}^{th}}\] by subtracting two.

Complete step by step answer:
As we can notice, we have our problem as an arithmetic sequence, we will apply concepts of arithmetic sequence.
An arithmetic sequence is a sequence (list of numbers) that has a common difference (a positive or negative constant) between the two consecutive terms.
According to our problem statement sequence, two consecutive terms always differ by $2$, which means that if we know the ${{n}^{th}}$ terms, we will get the \[n+{{1}^{th}}\] by subtracting two from the previous number.
We will start from ${{a}_{0}}=7$, which is the starting point of our arithmetic sequence.
The next term, that is ${{a}_{1}}$, will be \[{{a}_{0}}-2=7-2=5\], and so on.
The general rule states that we just have to described with words:
start from the initial value $7$, and subtract $2$ with each iteration.
This means that, after $n$ iterations, we will have subtracted two $n$ times, that is, we will have to subtract a total of $2n$.
Therefore, the rule of our arithmetic sequence \[7,5,3,1,-1\] will be:
${{a}_{n}}=7-2n$

Note: As this is an arithmetic sequence, we can confirm that our rule defined is correct or not.
We can confirm this by building some terms using the definition: given the starting value ${{a}_{0}}$,
we have
\[{{a}_{1}}={{a}_{0}}-1\cdot 2\]
\[{{a}_{2}}={{a}_{1}}-2=\left( {{a}_{0}}-2 \right)-2={{a}_{0}}-2\cdot 2\]
\[{{a}_{3}}={{a}_{2}}-2=({{a}_{0}}-2\cdot 2)-2={{a}_{0}}-3\cdot 2\]
\[{{a}_{4}}={{a}_{3}}-2=({{a}_{0}}-3\cdot 2)-2={{a}_{0}}-4\cdot 2\]
As we can see from above, the index of the term is equal to the times we have to subtract $2$.
Therefore, we can say that our solution (rule) is correct.