If `a_1, a_2, a_3,.......` are in G.P. then the value of determinant `|(log(a_n), log(a_(n+1)), log(a_(n+2))), (log(a_(n+3)), log(a_(n+4)), log(a_(n+5))), (log(a_(n+6)), log(a_(n+7)), log(a_(n+8)))|` equals (A) 0 (B) 1 (C) 2 (D) 3

Options

A. 1
B. 0
C. 2a
D. a

Answer

B

Explanation

we have
`(a_(n+1))^(2) =a_(n)a_(n)+2`
`rArr 2 log a_(n+1) =log a_(n)+ log a_(n+2)`
Similarly
`2 log a_(n+4) =log a_(n+3)+ log a_(n+2)`
`2log a_(n+7)= log a_(n+6) + log a+(n+8)`
Substituting these values in second column of determinant we get
`Delta =(1)/(2) |{:(log a_(n),,loga_(n)+loga_(n+2),,log a_(n+2)),(log a_(n+3),,log a_(n+3)+log a_(n+5),,log a_(n+5)),(log a_(n+6),,log_(n+6)+loga_(n+8),,log a_(n+8)):}|`
`=(1)/(2) (0) =0" ""[Using " C_(2) to C_(2) -C_(1)-C_(3)"]"`