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Ifn! $3 \times(n !)$ and $(n+1) !$ are in GP, then the value of $n$ will be
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The correct answer is:
8
Let n!, $3(\mathrm{n} !)$ and $(\mathrm{n}+1) !$ are in G. $P$. Then, $[3(\mathrm{n} !)]^{2}=(\mathrm{n} !)(\mathrm{n}+1) !$
$\Rightarrow 9 \times n ! \times n !=(n !)(n+1) n !$
$\Rightarrow 9=(\mathrm{n}+1) \Rightarrow \mathrm{n}=8$
$\Rightarrow 9 \times n ! \times n !=(n !)(n+1) n !$
$\Rightarrow 9=(\mathrm{n}+1) \Rightarrow \mathrm{n}=8$
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