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), thenA. `k=-1`B. `k=1`C. `k=0`D. None of these

Let `x gt 0`, `y gt 0`, `z gt 0` are respectively the `2^(nd)`, `3^(rd

1.	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), thenA. `k=-1`B. `k=1`C. `k=0`D. None of these
Answer» Correct Answer - A `(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)))`