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

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(C) The given lines are $L_1: x+2ay+a=0$,$L_2: x+3by+b=0$,and $L_3: x+4cy+c=0$. Since these lines are concurrent,the determinant of their coefficients

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harmonic progression

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(C) The given lines are $L_1: x+2ay+a=0$,$L_2: x+3by+b=0$,and $L_3: x+4cy+c=0$. Since these lines are concurrent,the determinant of their coefficients must be zero: $\left|\begin{array}{lll} 1 & 2a & a \ 1 & 3b & b \ 1 & 4c & c \end{array}ight|=0$. Applying row operations $R_1 ightarrow R_1-R_2$ and $R_2 ightarrow R_2-R_3$: $\left|\begin{array}{ccc} 0 & 2a-3b & a-b \ 0 & 3b-4c & b-c \ 1 & 4c & c \end{array}ight|=0$. Expanding along the first column: $(2a-3b)(b-c) - (3b-4c)(a-b) = 0$. $2ab - 2ac - 3b^2 + 3bc - (3ab - 3b^2 - 4ac + 4bc) = 0$. $2ab - 2ac - 3b^2 + 3bc - 3ab + 3b^2 + 4ac - 4bc = 0$. $-ab - bc + 2ac = 0$. $2ac = ab + bc$. Dividing by $abc$: $\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$. This is the condition for $a, b, c$ to be in harmonic progression.

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

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