Question

{"given":"First term $= x$, $(2n-1)$th term $= y$ for AP, GP, and HP. The $n$th term is the middle term between positions $1$ and $2n-1$.","key_observation":"The $n$th term of AP, GP, HP with equal first and last terms equals their AM, GM, and HM respectively. By the AM-GM-HM inequality: $a \\geq b \\geq c$, and $ac = b^2$ always holds.","option_analysis":[{"label":"(A)","text":"$a = b = c$","verdict":"correct","explanation":"Equality $a = b = c$ holds when $x = y$ (all terms of each progression are equal). Since the problem does not exclude this case, and the inequality $a \\geq b \\geq c$ includes equality, option A can be true."},{"label":"(B)","text":"$a \\geq b \\geq c$","verdict":"correct","explanation":"By the AM-GM-HM inequality, $a = \\frac{x+y}{2} \\geq \\sqrt{xy} = b \\geq \\frac{2xy}{x+y} = c$ always holds for positive $x, y$."},{"label":"(C)","text":"$a + c = b$","verdict":"incorrect","explanation":"We have $a = \\frac{x+y}{2}$, $b = \\sqrt{xy}$, $c = \\frac{2xy}{x+y}$; in general $a + c \\neq b$, and since $a \\geq b$ alone, $a + c > b$ always."},{"label":"(D)","text":"$ac - b^2 = 0$","verdict":"correct","explanation":"$$ac = \\frac{x+y}{2} \\cdot \\frac{2xy}{x+y} = xy = (\\sqrt{xy})^2 = b^2$$ so $ac - b^2 = 0$ always."}],"answer":"(A), (B), (D)","formula_steps":[]}