If `x_(1), x_(2), x_(3)` are the abcissa of the points `A_(1), A_(2), A_(3)` respectively where the lines `y=m_(1)x, y=m_(2)x, y =m_(3)` x meet the line `2x-y+3=0` such that `m_(1),m_(2),m_(3)` are in A.P., then `x_(1),x_(2),x_(3)` are in

Options

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

Answer

To solve the problem step by step, we will analyze the given lines and their intersections with the line \(2x - y + 3 = 0\).

### Step 1: Identify the equations of the lines
We have three lines given by:
1. \(y = m_1 x\)
2. \(y = m_2 x\)
3. \(y = m_3 x\)

And they intersect with the line:
\[ 2x - y + 3 = 0 \]
This can be rewritten as:
\[ y = 2x + 3 \]

### Step 2: Find the intersection points
To find the x-coordinates of the intersection points, we will substitute \(y\) from the line \(y = m_ix\) into the line \(y = 2x + 3\).

For the line \(y = m_1 x\):
\[ m_1 x = 2x + 3 \]
Rearranging gives:
\[ (m_1 - 2)x = 3 \implies x_1 = \frac{3}{m_1 - 2} \]

For the line \(y = m_2 x\):
\[ m_2 x = 2x + 3 \]
Rearranging gives:
\[ (m_2 - 2)x = 3 \implies x_2 = \frac{3}{m_2 - 2} \]

For the line \(y = m_3 x\):
\[ m_3 x = 2x + 3 \]
Rearranging gives:
\[ (m_3 - 2)x = 3 \implies x_3 = \frac{3}{m_3 - 2} \]

### Step 3: Analyze the slopes
We know that \(m_1, m_2, m_3\) are in Arithmetic Progression (A.P.). This means:
\[ 2m_2 = m_1 + m_3 \]

### Step 4: Transform the slopes
Subtracting 2 from each slope gives us:
- \(m_1 - 2\)
- \(m_2 - 2\)
- \(m_3 - 2\)

Since \(m_1, m_2, m_3\) are in A.P., \(m_1 - 2, m_2 - 2, m_3 - 2\) will also be in A.P.

### Step 5: Reciprocal transformation
The x-coordinates \(x_1, x_2, x_3\) can be expressed as:
\[ x_1 = \frac{3}{m_1 - 2}, \quad x_2 = \frac{3}{m_2 - 2}, \quad x_3 = \frac{3}{m_3 - 2} \]

Since \(m_1 - 2, m_2 - 2, m_3 - 2\) are in A.P., the reciprocals \( \frac{1}{m_1 - 2}, \frac{1}{m_2 - 2}, \frac{1}{m_3 - 2} \) will be in Harmonic Progression (H.P.).

### Step 6: Conclusion
Since \(x_1, x_2, x_3\) are the reciprocals of \(m_1 - 2, m_2 - 2, m_3 - 2\), and these are in H.P., it follows that \(x_1, x_2, x_3\) are in H.P. as well.

### Final Answer
Thus, \(x_1, x_2, x_3\) are in H.P.

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