MHTCET 22 Apr 2025 Shift 2 Paper

Concept:For three straight lines to be concurrent, the determinant of their coefficients must be zero.Explanation:The given lines are:$x + 2ay + a = 0$,$x + 3by + b = 0$,$x + 4cy + c = 0$.For concurrency, set the determinant of coefficients equal to zero:$$\begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$$Expanding along the first row:$1(3bc - 4bc) - 2a(c - b) + a(4c - 3b) = 0$Simplifying:$-bc - 2ac + 2ab + 4ac - 3ab = 0$$2ac - ab - bc = 0$Therefore:$2ac = ab + bc$Dividing by $abc$:$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$This is the condition for harmonic progression.Answer:$a, b, c$ are in Harmonic progression.Correct option: A. Harmonic progression.