First we need to find what kind of sequence we have. To do that we are going to test if the sequence as a difference [tex]d[/tex], in which case it will be an arithmetic sequence, or a ratio [tex]r[/tex], in which case it will be geometric sequence.
The formula to find [tex]d[/tex] is [tex]d=a_{n}-a_{n-a}[/tex]
where
[tex]a_{n}[/tex] is the current term in the sequence
[tex]a_{n-1}[/tex] is the previous term in the sequence
The formula to find [tex]r[/tex] is [tex]r= \frac{a _{n} }{a _{n-1} } [/tex]
where
[tex]a_{n}[/tex] is the current term in the sequence
[tex]a_{n-1}[/tex] is the previous term in the sequence
- For [tex]a_{n}=10[/tex] and [tex]a_{n-1}=7[/tex]
[tex]d=10-3[/tex]
[tex]d=3[/tex]
[tex]r= \frac{10}{7} [/tex]
- For [tex]a_{n}=4[/tex] and [tex]a_{n-1}=1[/tex]:
[tex]d=4-1[/tex]
[tex]d=3[/tex]
[tex]r= \frac{4}{1} [/tex]
[tex]r=4[/tex]
Since [tex]d[/tex] is equal in both procedures whereas [tex]r[/tex] is not, we can conclude that our sequence is an arithmetic sequence.
Now we are going to use the explicit formula of an arithmetic sequence [tex]a_{n}=a_{1}+(n-1)d[/tex]
where
[tex]a_{n}[/tex] is the nth term
[tex]a_{1}[/tex] is the first term
[tex]n[/tex] is the position of the term in the sequence
[tex]d[/tex] is the difference
We know that [tex]a_{1}=-2[/tex] and [tex]d=3[/tex], so lets replace those values in our formula:
[tex]a_{n}=-2+(n-1)(3)[/tex]
[tex]a_{n}=-2+3n-3[/tex]
[tex]a_{n}=3n-5[/tex]
We can conclude that the correct answer is D) an = 3n - 5
