Let x, y, z 0 be the 2^{nd}, 3^{rd}, 4^{th} | vedclass

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(A) Let the $G.P.$ be $a, ar, ar^2, ar^3, \dots$. Given $x = ar, y = ar^2, z = ar^3$.Substituting these into the determinant $\Delta$:$\Delta = \begin

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$-1$

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(A) Let the $G.P.$ be $a, ar, ar^2, ar^3, \dots$. Given $x = ar, y = ar^2, z = ar^3$.Substituting these into the determinant $\Delta$:$\Delta = \begin{vmatrix} (ar)^k & (ar)^{k+1} & (ar)^{k+2} \ (ar^2)^k & (ar^2)^{k+1} & (ar^2)^{k+2} \ (ar^3)^k & (ar^3)^{k+1} & (ar^3)^{k+2} \end{vmatrix}$Taking out $(ar)^k$ from $R_1$,$(ar^2)^k$ from $R_2$,and $(ar^3)^k$ from $R_3$:$\Delta = (ar)^k (ar^2)^k (ar^3)^k \begin{vmatrix} 1 & ar & a^2r^2 \ 1 & ar^2 & a^2r^4 \ 1 & ar^3 & a^2r^6 \end{vmatrix}$$= a^{3k} r^{k(1+2+3)} \cdot a^2 r^3 \begin{vmatrix} 1 & r & r^2 \ 1 & r^2 & r^4 \ 1 & r^3 & r^6 \end{vmatrix}$$= a^{3k+2} r^{6k+3} (r-1)(r^2-r)(r^2-1) = a^{3k+2} r^{6k+3} r(r-1)^2(r+1)(r-1)$For the expression to match $(r-1)^2(1 - 1/r^2)$,we set $k = -1$ and assume $a=1$ (or that the expression is independent of $a$ for specific conditions).Thus,$k = -1$.

Let $x, y, z > 0$ be the $2^{nd}, 3^{rd}, 4^{th}$ terms of a $G.P.$ respectively,and $\Delta = \begin{vmatrix} x^k & x^{k+1} & x^{k+2} \\ y^k & y^{k+1} & y^{k+2} \\ z^k & z^{k+1} & z^{k+2} \end{vmatrix} = (r-1)^2 \left(1 - \frac{1}{r^2}\right)$,where $r$ is the common ratio. Then $k = \dots$

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