Sequence And Series Question 180
Question: $ \alpha ,\ \beta $ are the roots of the equation $ x^{2}-3x+a=0 $ and $ \gamma ,\ \delta $ are the roots of the equation $ x^{2}-12x+b=0 $ . If $ \alpha ,\ \beta ,\ \gamma ,\ \delta $ form an increasing G.P., then $ (a,\ b)= $
[DCE 2000]
Options:
A) (3, 12)
B) (12, 3)
C) (2, 32)
D) (4, 16)
Show Answer
Answer:
Correct Answer: C
Solution:
Since $ \alpha ,\ \beta ,\ \gamma ,\ \delta $ form an increasing G.P., so $ \alpha \delta =\beta \gamma $ where $ \alpha <\beta <\gamma <\delta $ . On solving $ x^{2}-3x+a=0 $ , we get $ x=\frac{1}{2}(3\pm \sqrt{9-4a}) $ . Also $ \alpha <\beta $ . Hence $ \alpha =\frac{1}{2}(3-\sqrt{9-4a}),\ \beta =\frac{1}{2}(3+\sqrt{9-4a}) $ Similarly from $ x^{2}-12x+b=0 $ , we get $ \gamma =\frac{1}{2}(12-\sqrt{144-4b}),\ \delta =\frac{1}{2}(12+\sqrt{144-4b}) $ Substituting these values of $ \alpha ,\ \beta ,\ \gamma ,\ \delta $ in $ \alpha \delta =\beta \gamma $ and simplifying, we get $ (a,\ b)=(2,\ 32) $ . Trick: Check the alternates; only (c) satisfies the condition.
BETA
