Let `x gt 0`, `y gt 0`, `z gt 0` are respectively the `2^(nd)`, `3^(rd)`, `4^(th)` terms of a `G.P.`and `Delta=|{:(x^(k),x^(k+1),x^(k+2)),(y^(k),y^(k+1),y^(k+2)),(z^(k),z^(k+1),z^(k+2)):}|=(r-1)^(2)(1-(1)/(r^(2)))` (where `r` is the common ratio), then

Options

A. `k=-1`
B. `k=1`
C. `k=0`
D. None of these

Answer

A

Explanation

`(a)` `Delta=x^(k)y^(k)z^(k)|{:(1,ar,a^(2)r^(2)),(1,ar^(2),a^(2)r^(4)),(1,ar^(3),a^(2)r^(6)):}|`
`=a^(3k)*r^(6k)*a^(3)r^(3)|{:(1,1,1),(1,r,r^(2)),(1,r^(2),r^(4)):}|`
`=a^(3(k+1))*r^(6k+3)*(1-r)(r-r^(2))(r^(2)-1)`
Clearly, `k=-1`
`:. Delta=r^(-2)(1-r)^(2)(r^(2)-1)`
`=(r-1)^(2)(1-(1)/(r^(2)))`