MHT CET202522 Apr 2025Evening ShiftMathematicsDeterminantsActual
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 , c are in
Options
- AHarmonic progression
- BGeometric progression
- CArithmetic progression
- DArithmetico geometric progression
Correct answer
A. Harmonic progression
Step-by-step solution
The concurrency of lines x + 2ay + a = 0 , x + 3by + b = 0 , and x + 4cy + c = 0 requires the determinant of their coefficients to vanish: vmatrix 1 & 2a & a 1 & 3b & b 1 & 4c & c vmatrix = 0. Expanding this determinant yields -bc + 2ac - ab = 0 . For nonzero a , b , and c , dividing by abc gives - 1 a + 2 b - 1 c = 0, which simplifies to 2 b = 1 a + 1 c . This is the condition for 1 a , 1 b , 1 c to be in arithmetic progression, implying that a , b , and c are in harmonic progression. Answer: A