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Q.

If $x_{1}, x_{3}$ are the roots of $a x^{2} - 6 x + 1 = 0$ and $x_{2}, x_{4}$ are the roots of $21 x^{2} - b x + 1 = 0$ , If $x_{1}, x_{2}, x_{3}, x_{4}$ are in H.P, then $(a, b) =$

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a

$(5, 10)$

b

$(3, 8)$

c

$(- 5, 10)$

d

$(3, - 10)$

answer is C.

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Detailed Solution

$x_{1} + x_{3} = \frac{6}{a}, x_{1} x_{3} = \frac{1}{a} x_{2} + x_{4} = \frac{b}{21}, x_{2} x_{4} = \frac{1}{21} \frac{1}{x_{1}}, \frac{1}{x_{2}}, \frac{1}{x_{3}}, \frac{1}{x_{4}} a r e i n A . P \frac{1}{x_{1}} - \frac{1}{x_{3}} = \frac{1}{x_{2}} - \frac{1}{x_{4}} \Rightarrow \sqrt{36 - 4 a} = \sqrt{b^{2} - 84} \Rightarrow b^{2} = 120 - 4 a$

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