If $n!$ , $3 (n!)$ and $(n + 1)!$ are in G.P., then $n!$ , $5 (n!)$ and $(n + 1)!$ are in

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Solution

The correct option is A AP
$\frac{(n + 1)!}{3 (n!)} = \frac{3 (n!)}{n!}$
$⟹ \frac{3}{1} = \frac{n + 1}{3}$
$⟹ n = 8$
$8! = 40320$
$5 * 8! = 201600$
$9! = 362880$
Hence, they are in AP

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