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

Step-by-step solution

Let these three lines be L₁, L₂ and L₃ aligned & L₁=x+2 a y+a=0 & L₂=x+3 b y+b=0 & L₃=x+2 c y+c=0 aligned If L₁, L₂ and L₃ are concurrent, then | array lll 1 & 2 a & a 1 & 3 b & b 1 & 4 c & c array |=0 Applying R₁ R₁-R₂, R₂ R₂-R₃ , aligned & | array ccc 0 & 2 a-3 b & a-b 0 & 3 b-4 c & b-c 1 & 4 c & c array |=0 & 1[(2 a-3 b)(b-c)-(3 b-4 c)(a-b)]=0 & (2 a-3 b)(b-c)=(3 b-4 c)(a-b) & 2 a b-2 c a-3 b^2+3 b c=3 a b-3 b^2-4 c a+4 b c & a b+b c=2 c a & 1 a + 1 c = 2 b aligned Hence, a, b and c are in Harmonic progression.