If the lines $x+2 a y+a=0, x+3 b y+b=0$, $x+4 c y+c=0$ are concurrent, then $a, b$ and $c$ are in
If the lines $x+2 a y+a=0, x+3 b y+b=0$, $x+4 c y+c=0$ are concurrent, then $a, b$ and $c$ are in
arithmetic progression
geometric progression
harmonic progression
arithmetico-geometric progression
Solution
Let these three lines be $L_1, L_2$ and $L_3$
$\begin{aligned}
& L_1=x+2 a y+a=0 \\
& L_2=x+3 b y+b=0 \\
& L_3=x+2 c y+c=0
\end{aligned}$
If $L_1, L_2$ and $L_3$ are concurrent, then
$\left|\begin{array}{lll}
1 & 2 a & a \\
1 & 3 b & b \\
1 & 4 c & c
\end{array}\right|=0$
Applying $R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3$,
$\begin{aligned}
& \left|\begin{array}{ccc}
0 & 2 a-3 b & a-b \\
0 & 3 b-4 c & b-c \\
1 & 4 c & c
\end{array}\right|=0 \\
& \Rightarrow \quad 1[(2 a-3 b)(b-c)-(3 b-4 c)(a-b)]=0 \\
& \Rightarrow \quad(2 a-3 b)(b-c)=(3 b-4 c)(a-b) \\
& \Rightarrow \quad 2 a b-2 c a-3 b^2+3 b c=3 a b-3 b^2-4 c a+4 b c \\
& \Rightarrow \quad a b+b c=2 c a \\
& \Rightarrow \quad \frac{1}{a}+\frac{1}{c}=\frac{2}{b}
\end{aligned}$
Hence, $a, b$ and $c$ are in Harmonic progression.