Question

I need help solving thisIt’s from my prep guideI will add two additional pictures with the answer options

Answer 1

The formula for a general term in a geometric sequence is given by

`a_n=a_1r^(n-1)`

The first term of our sequence is 16, then, our geometric sequence would be

`a_n=16r^(n-1)`

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

`a_n=a_1+(n-1)d`

Where d is the common ratio.

Again, the first term is 16. Using the second term, we can find the value for d.

`\begin{gathered} a_n=16+(n-1)d \\ a_2=(15)/(2)\Rightarrow d=(15)/(2)-16=-(17)/(2) \end{gathered}`

This arithmetic sequence is

`a_n=16-(17)/(2)(n-1)`

If we evaluate bigger values for n to check with the other given terms, we have a match.

`\begin{gathered} a_3=16-(17)/(2)(3-1)=16-17=-1 \\ a_4=16-(17)/(2)(4-1)=(32)/(2)-(51)/(2)=-(19)/(2) \end{gathered}`

Thus, this series converge because it is arithmetic.