Question
Find k such that $\text{k}+9,\text{k}-6$ and 4 from three consecutive terms of a G.P.
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$\frac{n^2-9}{(n+3) !}+\frac{6}{(n+2) !}-\frac{1}{(n+1) !}$
$3, \frac{6}{5}, \frac{9}{25}, \frac{12}{125}, \frac{15}{625}, \ldots$