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

Explanation

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Identify the terms of the G.P.
Given that \( x, y, z \) are the 2nd, 3rd, and 4th terms of a geometric progression (G.P.), we can express them in terms of the first term \( a \) and the common ratio \( r \):
- \( x = ar \) (2nd term)
- \( y = ar^2 \) (3rd term)
- \( z = ar^3 \) (4th term)

### Step 2: Write the determinant
The determinant is given as:
\[ \Delta = \begin{vmatrix} x^k & x^{k+1} & x^{k+2} \\ y^k & y^{k+1} & y^{k+2} \\ z^k & z^{k+1} & z^{k+2} \end{vmatrix} \]

### Step 3: Factor out common terms
We can factor out \( x^k, y^k, z^k \) from each row:
\[ \Delta = x^k y^k z^k \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \]

### Step 4: Substitute the values of \( x, y, z \)
Substituting \( x, y, z \) in the determinant:
\[ \Delta = (ar)^k (ar^2)^k (ar^3)^k \begin{vmatrix} 1 & ar & (ar)^2 \\ 1 & ar^2 & (ar^2)^2 \\ 1 & ar^3 & (ar^3)^2 \end{vmatrix} \]
This simplifies to:
\[ \Delta = a^{3k} r^{6k} \begin{vmatrix} 1 & ar & a^2 r^2 \\ 1 & ar^2 & a^2 r^4 \\ 1 & ar^3 & a^2 r^6 \end{vmatrix} \]

### Step 5: Simplify the determinant
Now, we can factor \( a^2 \) from the second and third columns:
\[ \Delta = a^{3k + 2} r^{6k} a^2 \begin{vmatrix} 1 & r & r^2 \\ 1 & r^2 & r^4 \\ 1 & r^3 & r^6 \end{vmatrix} \]
This gives:
\[ \Delta = a^{3k + 2} r^{6k} \cdot a^2 \cdot \begin{vmatrix} 1 & r & r^2 \\ 1 & r^2 & r^4 \\ 1 & r^3 & r^6 \end{vmatrix} \]

### Step 6: Calculate the determinant
Now, we can compute the determinant:
\[ \begin{vmatrix} 1 & r & r^2 \\ 1 & r^2 & r^4 \\ 1 & r^3 & r^6 \end{vmatrix} = (r - 1)(r^2 - r)(r^3 - r^2) = (r - 1)(r^2 - r)(r^3 - r^2) \]
This determinant simplifies to \( (r - 1)(r^2 - 1)(r^3 - 1) \).

### Step 7: Set the determinant equal to the given expression
According to the problem, we have:
\[ \Delta = (r - 1)^2 \left(1 - \frac{1}{r^2}\right) \]
Setting the two expressions for \( \Delta \) equal gives:
\[ a^{3k + 2} r^{6k} \cdot (r - 1)(r^2 - 1)(r^3 - 1) = (r - 1)^2 \left(1 - \frac{1}{r^2}\right) \]

### Step 8: Equate powers of \( a \)
Since the left-hand side contains \( a \) and the right-hand side does not, we set the exponent of \( a \) to zero:
\[ 3k + 2 = 0 \implies k = -\frac{2}{3} \]

### Step 9: Final result
Thus, the value of \( k \) is:
\[ \boxed{-1} \]