Question:
The lines $x + 2ay + \text{a} = 0, x + 3by + \text{b} = 0, x + 4cy + \text{c} = 0$ are concurrent then $a, b, c$ are in}
The lines $x + 2ay + \text{a} = 0, x + 3by + \text{b} = 0, x + 4cy + \text{c} = 0$ are concurrent then $a, b, c$ are in}
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Harmonic Mean $b = \frac{2ac}{a+c} \iff$ H.P.
Updated On: Apr 26, 2026
- Harmonic progression
- Geometric progression
- Arithmetic progression
- Arithmetico geometric progression
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The Correct Option is A
Solution and Explanation
Step 1: Determinant for Concurrency
Lines $L_1, L_2, L_3$ are concurrent if $\begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$.
Step 2: Expand Determinant
$1(3bc - 4bc) - 2a(c - b) + a(4c - 3b) = 0$
$-bc - 2ac + 2ab + 4ac - 3ab = 0$
$2ac - ab - bc = 0 \implies 2ac = b(a+c)$.
Step 3: Conclusion
$b = \frac{2ac}{a+c}$. This is the condition for $a, b, c$ to be in Harmonic Progression (H.P.).
Final Answer: (A)
Lines $L_1, L_2, L_3$ are concurrent if $\begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$.
Step 2: Expand Determinant
$1(3bc - 4bc) - 2a(c - b) + a(4c - 3b) = 0$
$-bc - 2ac + 2ab + 4ac - 3ab = 0$
$2ac - ab - bc = 0 \implies 2ac = b(a+c)$.
Step 3: Conclusion
$b = \frac{2ac}{a+c}$. This is the condition for $a, b, c$ to be in Harmonic Progression (H.P.).
Final Answer: (A)
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