If x_1, x_3 are the roots of A x^2 - 4 x + 1 | vedclass

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(A) Given that $x_1, x_2, x_3, x_4$ are in harmonic progression $(HP)$,their reciprocals $\frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}, \frac{1}{x_4}$

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$\frac{11}{5}$

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(A) Given that $x_1, x_2, x_3, x_4$ are in harmonic progression $(HP)$,their reciprocals $\frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}, \frac{1}{x_4}$ are in arithmetic progression $(AP)$. Let these be $a-3d, a-d, a+d, a+3d$. From $A x^2 - 4 x + 1 = 0$,the roots $x_1, x_3$ satisfy $\frac{1}{x_1} + \frac{1}{x_3} = 4$ and $\frac{1}{x_1} \cdot \frac{1}{x_3} = A$. Substituting the $AP$ terms: $(a-3d) + (a+d) = 4 \implies 2a - 2d = 4 \implies a - d = 2$. Also,$(a-3d)(a+d) = A$. From $B x^2 - 6 x + 1 = 0$,the roots $x_2, x_4$ satisfy $\frac{1}{x_2} + \frac{1}{x_4} = 6$ and $\frac{1}{x_2} \cdot \frac{1}{x_4} = B$. Substituting the $AP$ terms: $(a-d) + (a+3d) = 6 \implies 2a + 2d = 6 \implies a + d = 3$. Solving the system $a-d=2$ and $a+d=3$,we get $2a=5 \implies a=2.5$ and $2d=1 \implies d=0.5$. Now,$A = (a-3d)(a+d) = (2.5 - 1.5)(3) = 1 	imes 3 = 3$. And $B = (a-d)(a+3d) = (2)(2.5 + 1.5) = 2 	imes 4 = 8$. Finally,$\frac{B+A}{B-A} = \frac{8+3}{8-3} = \frac{11}{5}$.

If $x_1, x_3$ are the roots of $A x^2 - 4 x + 1 = 0$ and $x_2, x_4$ are the roots of $B x^2 - 6 x + 1 = 0$ such that $x_1, x_2, x_3, x_4$ are in harmonic progression,then $\frac{B+A}{B-A} = $

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