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 prove that the determinant \( \Delta \) is equal to zero given that \( a_1, a_2, a_3 \) are in arithmetic progression (A.P.) and \( b_1, b_2, b_3 \) are in harmonic progression (H.P.), we will follow these steps:

### Step 1: Express \( a_2 \) and \( a_3 \) in terms of \( a_1 \)
Since \( a_1, a_2, a_3 \) are in A.P., we can express them as:
- \( a_2 = a_1 + d \)
- \( a_3 = a_1 + 2d \)

where \( d \) is the common difference.

### Step 2: Express \( b_1, b_2, b_3 \) in terms of a common variable
Since \( b_1, b_2, b_3 \) are in 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:
- \( x_2 = x_1 + d' \)
- \( x_3 = x_1 + 2d' \)

for some common difference \( d' \).

### Step 3: Set up the determinant \( \Delta \)
We need to compute the determinant:
\[ \Delta = \begin{vmatrix} a_1 - b_1 & a_1 - b_2 & a_1 - b_3 \\ a_2 - b_1 & a_2 - b_2 & a_2 - b_3 \\ a_3 - b_1 & a_3 - b_2 & a_3 - b_3 \end{vmatrix} \]

Substituting \( a_2 \) and \( a_3 \):
\[ \Delta = \begin{vmatrix} a_1 - b_1 & a_1 - b_2 & a_1 - b_3 \\ (a_1 + d) - b_1 & (a_1 + d) - b_2 & (a_1 + d) - b_3 \\ (a_1 + 2d) - b_1 & (a_1 + 2d) - b_2 & (a_1 + 2d) - b_3 \end{vmatrix} \]

### Step 4: Simplify the determinant
Now, we will perform column operations. 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 5: Factor out common terms
Notice that the second and third rows are proportional:
\[ \Delta = d \cdot \begin{vmatrix} a_1 - b_1 & b_1 - b_2 & b_1 - b_3 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{vmatrix} \]

### Step 6: Evaluate the determinant
Since the second and third rows are identical (both are multiples of each other), the determinant evaluates to zero:
\[ \Delta = 0 \]

### Conclusion
Thus, we have shown that \( \Delta = 0 \).