If `n!,3xxn! and (n+1)!` are in GP, then `n!,5xxn! And (n+1)!` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

A

Explanation

It is given that `n!,3xxn!and(n+1)!` are in G.P.
`:.(3xxn!)^(2)=n!xx(n+1)!`
`implies9xx(n!)^(2)=n!xx(n+1)n!`
`implies9=n+1`
`implies` n = 8
`:.n!""=8!,5xxn!""5=5xx8!and(n+1)!""=9!`
We observe that
`2(5xxn!)=(9+1)xx8!""=9xx8!+8!`
`implies2(5xxn!)=9!+8!""=n!+(n+1)!`
Hence, `n!5xxn!and(n+1)!` are in A.P.