Find k so that the numbers 2k – 1, 4k + 1, and 8k + 11 form a geometric sequence.
Answer:
[tex]\boxed{k=2}[/tex]
Step-by-step explanation:
[tex]\text{For all geometric sequences, the following is true for consecutive terms $a, b, c$ : $b^2=ac$}\\\text{Therefore, we can construct this equation: $\boxed{(4k+1)^2=(2k-1)(8k+11)}$}\\\text{Solving it will give $\boxed{k=2}.$}[/tex]
