If the lines `a x+2y+1=0,b x+3y+1=0a n dc x+4y+1=0` are concurrent, then `a ,b ,c` are in (a). A.P. (b). G.P. (c). H.P. (d). none of these

Options

A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer

To determine the relationship between \( a, b, c \) when the lines \( ax + 2y + 1 = 0 \), \( bx + 3y + 1 = 0 \), and \( cx + 4y + 1 = 0 \) are concurrent, we can use the concept of determinants. The lines are concurrent if the determinant of their coefficients is equal to zero.

### Step-by-step Solution:

1. **Set up the determinant**:
The coefficients of the lines can be arranged in a determinant as follows:
\[ \begin{vmatrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{vmatrix} \]

2. **Calculate the determinant**:
The determinant can be calculated using the formula:
\[ D = a \begin{vmatrix} 3 & 1 \\ 4 & 1 \end{vmatrix} - 2 \begin{vmatrix} b & 1 \\ c & 1 \end{vmatrix} + 1 \begin{vmatrix} b & 3 \\ c & 4 \end{vmatrix} \]

- Calculate \( \begin{vmatrix} 3 & 1 \\ 4 & 1 \end{vmatrix} = (3 \cdot 1) - (4 \cdot 1) = 3 - 4 = -1 \)
- Calculate \( \begin{vmatrix} b & 1 \\ c & 1 \end{vmatrix} = (b \cdot 1) - (c \cdot 1) = b - c \)
- Calculate \( \begin{vmatrix} b & 3 \\ c & 4 \end{vmatrix} = (b \cdot 4) - (c \cdot 3) = 4b - 3c \)

Substituting these values into the determinant:
\[ D = a(-1) - 2(b - c) + (4b - 3c) \]
Simplifying this gives:
\[ D = -a - 2b + 2c + 4b - 3c = -a + 2b - c \]

3. **Set the determinant to zero**:
For the lines to be concurrent, we set the determinant equal to zero:
\[ -a + 2b - c = 0 \]
Rearranging this gives:
\[ 2b = a + c \]

4. **Identify the relationship**:
The equation \( 2b = a + c \) indicates that \( a, b, c \) are in Arithmetic Progression (A.P.). This is because the middle term \( b \) is the average of \( a \) and \( c \).

### Conclusion:
Thus, the correct answer is that \( a, b, c \) are in **A.P.**.