If the system of equations `x + 4ay + az = 0 and x + 3by + bz = 0 and x + 2cy + cz = 0` have a nonzero solution, then `a, b, c ( != 0)` are in

Options

A. AP
B. GP
C. HP
D. AGP

Answer

To solve the problem, we need to analyze the given system of equations and determine the conditions under which they have a non-zero solution. The equations are:

1. \( x + 4ay + az = 0 \)
2. \( x + 3by + bz = 0 \)
3. \( x + 2cy + cz = 0 \)

### Step 1: Write the system in matrix form

We can express the system of equations in the form of a matrix:

\[ \begin{bmatrix} 1 & 4a & a \\ 1 & 3b & b \\ 1 & 2c & c \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Set up the determinant

For the system to have a non-zero solution, the determinant of the coefficient matrix must be zero. Let's denote the determinant as \( D \):

\[ D = \begin{vmatrix} 1 & 4a & a \\ 1 & 3b & b \\ 1 & 2c & c \end{vmatrix} \]

### Step 3: Calculate the determinant

Using the determinant formula for a 3x3 matrix, we can expand \( D \):

\[ D = 1 \cdot \begin{vmatrix} 3b & b \\ 2c & c \end{vmatrix} - 4a \cdot \begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix} + a \cdot \begin{vmatrix} 1 & 3b \\ 1 & 2c \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 3b & b \\ 2c & c \end{vmatrix} = 3bc - 2bc = bc \)

2. \( \begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix} = c - b \)

3. \( \begin{vmatrix} 1 & 3b \\ 1 & 2c \end{vmatrix} = 2c - 3b \)

Substituting these back into the determinant expression:

\[ D = 1 \cdot (bc) - 4a(c - b) + a(2c - 3b) \]

### Step 4: Simplify the determinant

Now we simplify \( D \):

\[ D = bc - 4a(c - b) + a(2c - 3b) \]

Distributing the terms:

\[ D = bc - 4ac + 4ab + 2ac - 3ab \]

Combining like terms:

\[ D = bc - 4ac + (4ab - 3ab) + 2ac = bc - 2ac + ab \]

### Step 5: Set the determinant to zero

For the system to have a non-zero solution, we set \( D = 0 \):

\[ bc - 2ac + ab = 0 \]

### Step 6: Rearranging the equation

Rearranging gives us:

\[ ab + bc - 2ac = 0 \]

### Step 7: Factor the equation

We can factor the equation as follows:

\[ ab + bc = 2ac \]

### Step 8: Conclusion about \( a, b, c \)

This implies that \( \frac{1}{a} + \frac{1}{c} = \frac{2}{b} \), which is a condition for \( a, b, c \) to be in Harmonic Progression (HP). Thus, we conclude that \( a, b, c \) are in HP.