Question

If the lines \( x + 2ay + a = 0 \), \( x + 3by + b = 0 \), \( x + 4cy + c = 0 \) are concurrent, then \( a, b, c \) are in

• A. Arithmetic Progression
• B. Geometric Progression
• C. Harmonic Progression
• D. Arithmetico-geometric Progression

Hint:

To check concurrency of three lines, eliminate variables using pairwise subtraction and compare the resulting expressions to identify relationships among constants.

Answer 1

The Correct Option is C

Step 1: Understand the condition for concurrency 
If three lines are concurrent, their point of intersection \((x, y)\) satisfies all three equations: 

Step 2: Eliminate \( x \) 
Subtract (1) from (2): 
\[ (x + 3by + b) - (x + 2ay + a) = 0 \Rightarrow y(3b - 2a) + (b - a) = 0 \tag{4} \] Subtract (2) from (3): 
\[ (x + 4cy + c) - (x + 3by + b) = 0 \Rightarrow y(4c - 3b) + (c - b) = 0 \tag{5} \] Step 3: Solve the system of equations (4) and (5) 
From (4), solve for \( y \): 
\[ y = \frac{a - b}{3b - 2a} \] From (5), solve for \( y \): 
\[ y = \frac{b - c}{4c - 3b} \] Step 4: Equating both expressions for \( y \) 
\[ \frac{a - b}{3b - 2a} = \frac{b - c}{4c - 3b} \Rightarrow \text{Cross-multiply:} \] \[ (a - b)(4c - 3b) = (b - c)(3b - 2a) \] After simplification (expand both sides and compare), we get: 
\[ \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in A.P.} \Rightarrow a, b, c \text{ are in H.P.} \] % Final Result 
Hence, 
\[ \boxed{\text{Harmonic Progression}} \]