If positive numbers $a,b,c$ be in H.P, then the equation $x^2-kx+2b^{101}-a^{101}-c^{101}=0$ $(k\in R)$ has
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Both roots positive
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Both roots negative
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One positive and one negative root
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Both roots imaginary
Since a, b, c are in H.P., 2/b = 1/a + 1/c, which implies b = 2ac/(a+c). The constant term of the quadratic is 2b^101 - a^101 - c^101. By AM-GM inequality, b^101 < (a^101 + c^101)/2, so 2b^101 < a^101 + c^101, making the constant term negative. A quadratic with a negative constant term always has roots of opposite signs.
Since a, b, and c are in harmonic progression, their reciprocals are in arithmetic progression, meaning b = 2ac/(a+c). Because a and c are positive, this relation guarantees that 2b^101 - a^101 - c^101 is strictly less than zero. We can prove this by noting that for positive a and c, (a^101 + c^101)/2 is strictly greater than ((a+c)/2)^101, which in turn is greater than b^101, making the constant term of the quadratic negative. A negative constant term means the product of the roots is negative, ensuring the quadratic equation has one positive and one negative root. The result is one positive and one negative root.