Suppose `a_(1), a_(2)`, .... Are real numbers, with `a_(1) ne 0`. If `a_(1), a_(2), a_(3)`, ... Are in A.P., then

Options

A. `A=[(a_(1),a_(2),a_(3)),(a_(4),a_(5),a_(6)),(a_(5),a_(6),a_(7))]` is singular (where `i=sqrt(-1)`)
B. the system of equations `a_(1)x+a_(2)y+a_(3)z=0, a_(4)x+a_(5)y+a_(6)z=0, a_(7)x+a_(8)y+a_(9)z=0` has infinite number of solutions
C. `B[(a_(1),ia_(2)),(ia_(2),a_(1))]` is nonsingular
D. none of these

Explanation

To solve the problem, we need to analyze the conditions given in the question regarding the sequence of numbers \( a_1, a_2, a_3, \ldots \) being in Arithmetic Progression (A.P.) and the implications for the matrices formed by these numbers.

### Step-by-Step Solution:

1. **Understanding A.P.**:
Since \( a_1, a_2, a_3, \ldots \) are in A.P., we can express this relationship mathematically. For any three consecutive terms in A.P., the middle term is the average of the other two:
\[ a_2 = \frac{a_1 + a_3}{2} \implies a_3 = 2a_2 - a_1 \]

2. **Constructing the Matrix**:
We construct a matrix \( A \) using the terms \( a_1, a_2, a_3, a_4, a_5, a_6, a_7 \):
\[ A = \begin{bmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{bmatrix} \]

3. **Finding the Determinant**:
To check if the matrix \( A \) is singular (determinant = 0), we can perform row operations to simplify the determinant calculation.

We can transform the second and third rows:
\[ R_2 \rightarrow R_2 - R_1 \quad \text{and} \quad R_3 \rightarrow R_3 - R_2 \]
This gives us:
\[ A = \begin{bmatrix} a_1 & a_2 & a_3 \\ a_4 - a_1 & a_5 - a_2 & a_6 - a_3 \\ a_7 - a_4 & a_8 - a_5 & a_9 - a_6 \end{bmatrix} \]

4. **Using the A.P. Property**:
Since \( a_1, a_2, a_3 \) are in A.P., we can express \( a_4, a_5, a_6 \) in terms of \( a_1, a_2, a_3 \):
\[ a_4 = a_1 + d, \quad a_5 = a_2 + d, \quad a_6 = a_3 + d \]
where \( d \) is the common difference.

5. **Substituting Back**:
Substitute \( a_4, a_5, a_6 \) back into the determinant:
\[ A = \begin{bmatrix} a_1 & a_2 & a_3 \\ d & d & d \\ a_7 - (a_1 + d) & a_8 - (a_2 + d) & a_9 - (a_3 + d) \end{bmatrix} \]

6. **Determining Singularity**:
If \( a_4, a_5, a_6 \) are all equal to \( d \), then the second row becomes identical, leading to a determinant of zero. Thus, the matrix is singular.

7. **Conclusion**:
Therefore, we conclude that the matrix \( A \) is singular, which implies that the system of equations represented by this matrix has an infinite number of solutions.