The lines x+2ay+a=0,x+3by+b=0,and x+4cy+c=0 a | vedclass

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(A) Three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ are concurrent if the determinant of their coefficients is

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Harmonic progression

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(A) Three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ are concurrent if the determinant of their coefficients is zero:$\begin{vmatrix} 1 & 2a & a \ 1 & 3b & b \ 1 & 4c & c \end{vmatrix} = 0$Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\begin{vmatrix} 1 & 2a & a \ 0 & 3b-2a & b-a \ 0 & 4c-2a & c-a \end{vmatrix} = 0$Expanding along the first column:$(3b-2a)(c-a) - (b-a)(4c-2a) = 0$$3bc - 3ab - 2ac + 2a^2 - (4bc - 2ab - 4ac + 2a^2) = 0$$3bc - 3ab - 2ac + 2a^2 - 4bc + 2ab + 4ac - 2a^2 = 0$$-bc - ab + 2ac = 0$$2ac = ab + bc$Dividing both sides by $abc$:$\frac{2ac}{abc} = \frac{ab}{abc} + \frac{bc}{abc}$$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$This implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in arithmetic progression,which means $a, b, c$ are in harmonic progression.

The lines $x+2ay+a=0$,$x+3by+b=0$,and $x+4cy+c=0$ are concurrent. Then $a, b, c$ are in:

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