The formula for a general term in a geometric sequence is given by
[tex]a_n=a_1r^{n-1}[/tex]
The first term of our sequence is 16, then, our geometric sequence would be
[tex]a_n=16r^{n-1}[/tex]
Using the following terms, there's no r that fits for all terms, therefore, this sequence is not geometric.
The general term in a arithmetic sequence is
[tex]a_n=a_1+(n-1)d[/tex]
Where d is the common ratio.
Again, the first term is 16. Using the second term, we can find the value for d.
[tex]\begin{gathered} a_n=16+(n-1)d \\ a_2=\frac{15}{2}\Rightarrow d=\frac{15}{2}-16=-\frac{17}{2} \end{gathered}[/tex]
This arithmetic sequence is
[tex]a_n=16-\frac{17}{2}(n-1)[/tex]
If we evaluate bigger values for n to check with the other given terms, we have a match.
[tex]\begin{gathered} a_3=16-\frac{17}{2}(3-1)=16-17=-1 \\ a_4=16-\frac{17}{2}(4-1)=\frac{32}{2}-\frac{51}{2}=-\frac{19}{2} \end{gathered}[/tex]
Thus, this series converge because it is arithmetic.
