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  • 4 years ago

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

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  1. Mertsj
    • 4 years ago
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    \[\frac{y}{y-3}=\frac{3y+4}{y}\] \[y^2=3y^2-5y-12\] \[2y^2-5y-12=0\] \[(2y+3)(y-4)=0\] \[y=\frac{-3}{2} or y = 4\]

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