Given:
The sequence is,
[tex]k-3,k+2,k+3[/tex]
Suppose that the given sequence is in the geometric sequence. The common ratio will be the same.
[tex]\begin{gathered} \frac{k+2}{k-3}=\frac{k+3}{k+2} \\ (k+2)(k+2)=(k+3)(k-3) \\ k^2+4k+4=k^2-9 \\ 4k=-9-4 \\ 4k=-13 \\ k=-\frac{13}{4} \end{gathered}[/tex]
Answer: option c) -13/4
