Question

If $n!$, $3\times n!$ and $(n+1)!$ are in GP then $n!$, $5\times n!$ and $(n+1)!$ are in

• A. AP
• B. GP
• C. HP
• D. none of these

Answer 1

The correct option is A AP

Given
$(3n!{)}^2=(n+1)!n!$
$\Rightarrow 9=n+1\Rightarrow n=8$
$\therefore n!,5n!$ and $(n+1)!$ are $8!,5\times 8!$ and $9!$
Clearly, $9!+8!=10\times 8!=2(5\times 8!)$
$\therefore$ Numbers are in AP