(a) Simplify : \(\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\). (b) If \(\frac{1}{2}, \frac{1}{x}, \frac{1}{3}\) are successive terms of an arithmetic progressi...
Question 1 Report
(a) Simplify : \(\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\).
(b) If \(\frac{1}{2}, \frac{1}{x}, \frac{1}{3}\) are successive terms of an arithmetic progression (A.P), show that \(\frac{2 - x}{x - 3} = \frac{2}{3}\).
Answer Details
(a) Factorise numerator and denominator: \[x^2 - 8x + 16 = (x-4)^2,\qquad x^2 - 7x + 12 = (x-3)(x-4).\] \[\frac{(x-4)^2}{(x-3)(x-4)} = \frac{x-4}{x-3}\quad (x\neq 4).\]
(b) If \(\tfrac{1}{2}, \tfrac{1}{x}, \tfrac{1}{3}\) are successive terms of an A.P., the common difference is constant: \[\frac{1}{x} - \frac{1}{2} = \frac{1}{3} - \frac{1}{x}.\] The left side is \(\dfrac{2 - x}{2x}\) and the right side is \(\dfrac{x - 3}{3x}\), so \[\frac{2 - x}{2x} = \frac{x - 3}{3x}.\] Multiplying both sides by \(3x\) and dividing by \((x-3)\): \[\frac{2 - x}{x - 3} = \frac{2x}{3x} = \frac{2}{3}.\] Hence \(\dfrac{2 - x}{x - 3} = \dfrac{2}{3}\), as required.
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