If `a_(1),a_(2),a_(3),….,a_(r)` are in GP, then prove that the determinant `|(a_(r+1),a_(r+5),a_(r+9)),(a_(r+7),a_(r+11),a_(4+15)),(a_(r+11),a_(r+17),a_(r+21))|` is independent of `r`.

Explanation

To prove that the determinant

\[ D = \begin{vmatrix} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{vmatrix} \]

is independent of \( r \) when \( a_1, a_2, a_3, \ldots, a_r \) are in a geometric progression (GP), we will follow these steps:

### Step 1: Express terms in terms of the first term and common ratio
Assume the first term of the GP is \( A \) and the common ratio is \( R \). The \( n \)-th term of a GP can be expressed as:

\[ a_n = A R^{n-1} \]

Using this, we can express the terms in the determinant:

- \( a_{r+1} = A R^r \)
- \( a_{r+5} = A R^{r+4} \)
- \( a_{r+9} = A R^{r+8} \)
- \( a_{r+7} = A R^{r+6} \)
- \( a_{r+11} = A R^{r+10} \)
- \( a_{r+15} = A R^{r+14} \)
- \( a_{r+11} = A R^{r+10} \)
- \( a_{r+17} = A R^{r+16} \)
- \( a_{r+21} = A R^{r+20} \)

### Step 2: Substitute these expressions into the determinant
Now substitute these expressions into the determinant \( D \):

\[ D = \begin{vmatrix} A R^r & A R^{r+4} & A R^{r+8} \\ A R^{r+6} & A R^{r+10} & A R^{r+14} \\ A R^{r+10} & A R^{r+16} & A R^{r+20} \end{vmatrix} \]

### Step 3: Factor out the common terms
We can factor out \( A \) from each column:

\[ D = A^3 \begin{vmatrix} R^r & R^{r+4} & R^{r+8} \\ R^{r+6} & R^{r+10} & R^{r+14} \\ R^{r+10} & R^{r+16} & R^{r+20} \end{vmatrix} \]

### Step 4: Simplify the determinant
Next, we can factor out \( R^r \) from the first column, \( R^{r+6} \) from the second column, and \( R^{r+10} \) from the third column:

\[ D = A^3 R^{r + (r+6) + (r+10)} \begin{vmatrix} 1 & R^4 & R^8 \\ R^6 & R^{10} & R^{14} \\ R^{10} & R^{16} & R^{20} \end{vmatrix} \]

This simplifies to:

\[ D = A^3 R^{3r + 20} \begin{vmatrix} 1 & R^4 & R^8 \\ R^6 & R^{10} & R^{14} \\ R^{10} & R^{16} & R^{20} \end{vmatrix} \]

### Step 5: Evaluate the determinant
Now, we can evaluate the determinant:

\[ \begin{vmatrix} 1 & R^4 & R^8 \\ R^6 & R^{10} & R^{14} \\ R^{10} & R^{16} & R^{20} \end{vmatrix} \]

Notice that the second and third rows can be expressed as multiples of the first row, indicating that the determinant will be zero:

\[ D = 0 \]

### Conclusion
Since we have shown that the determinant \( D \) evaluates to zero, it is independent of \( r \).

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