Suppose `a_(1),a_(2),a_(3)` are in A.P. and `b_(1),b_(2),b_(3)` are in H.P. and let `Delta=|(a_(1)-b_(1),a_(1)-b_(2),a_(1)-b_(3)),(a_(2)-b_(1),a_(2)-b_(2),a_(2)-b_(3)),(a_(3)-b_(1),a_(3)-b_(2),a_(3)-b_(3))|` then prove that

Options

A. `Delta` is independent of `a_(1),a_(2),a_(3)`
B. `A_(1)-Delta,a_(2)-2Delta,a_(3)-3Delta` are in A.P.
C. `b_(1)+Delta,b_(2)+Delta^(2),b_(3)+Delta` are in H.P.
D. `Delta` is independent of `b_(1),b_(2),b_(3)`

Answer

To solve the given problem, we need to analyze the determinant defined by:

\[ \Delta = \begin{vmatrix} a_1 - b_1 & a_1 - b_2 & a_1 - b_3 \\ a_2 - b_1 & a_2 - b_2 & a_2 - b_3 \\ a_3 - b_1 & a_3 - b_2 & a_3 - b_3 \end{vmatrix} \]

### Step 1: Understand the sequences
Since \(a_1, a_2, a_3\) are in Arithmetic Progression (A.P.), we can express them as:
- \(a_2 = a_1 + d\)
- \(a_3 = a_1 + 2d\)

Where \(d\) is the common difference.

Since \(b_1, b_2, b_3\) are in Harmonic Progression (H.P.), we can express their reciprocals as being in A.P.:
- Let \(b_1 = \frac{1}{x_1}, b_2 = \frac{1}{x_2}, b_3 = \frac{1}{x_3}\) where \(x_1, x_2, x_3\) are in A.P.

Thus, we can write:
- \(x_2 = x_1 + d'\)
- \(x_3 = x_1 + 2d'\)

### Step 2: Substitute into the determinant
Now, substituting \(a_2\) and \(a_3\) into the determinant:

\[ \Delta = \begin{vmatrix} a_1 - b_1 & a_1 - b_2 & a_1 - b_3 \\ (a_1 + d) - b_1 & (a_1 + d) - b_2 & (a_1 + d) - b_3 \\ (a_1 + 2d) - b_1 & (a_1 + 2d) - b_2 & (a_1 + 2d) - b_3 \end{vmatrix} \]

### Step 3: Simplify the determinant
Now, we can perform column operations to simplify the determinant. We will subtract the first column from the second and third columns:

\[ \Delta = \begin{vmatrix} a_1 - b_1 & (a_1 - b_2) - (a_1 - b_1) & (a_1 - b_3) - (a_1 - b_1) \\ (a_1 + d) - b_1 & ((a_1 + d) - b_2) - ((a_1 + d) - b_1) & ((a_1 + d) - b_3) - ((a_1 + d) - b_1) \\ (a_1 + 2d) - b_1 & ((a_1 + 2d) - b_2) - ((a_1 + 2d) - b_1) & ((a_1 + 2d) - b_3) - ((a_1 + 2d) - b_1) \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} a_1 - b_1 & b_1 - b_2 & b_1 - b_3 \\ d & d & d \\ 2d & 2d & 2d \end{vmatrix} \]

### Step 4: Factor out common terms
Now, we can factor out the common terms from the second and third rows:

\[ \Delta = d \cdot 2d \cdot \begin{vmatrix} a_1 - b_1 & b_1 - b_2 & b_1 - b_3 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 5: Evaluate the determinant
The determinant of a matrix where two rows are identical is zero:

\[ \Delta = 0 \]

### Conclusion
Thus, we have shown that \(\Delta = 0\), which implies that the rows of the determinant are linearly dependent. Therefore, the conditions given in the problem are satisfied.