Question

If y-3, y, and 3y+4 are consecutive terms in a geometric sequence, determine the value(s) of y.

Answer 1

\[\frac{y}{y-3}=\frac{3y+4}{y}\]
\[y^2=3y^2-5y-12\]
\[2y^2-5y-12=0\]
\[(2y+3)(y-4)=0\]
\[y=\frac{-3}{2} or y = 4\]