Sequence And Series Question 344
Question: Given that $ \alpha ,,\gamma $ are root of the equation $ Ax^{2}-4x+1=0 $ . and $ \beta ,,\delta $ the roots of the equation $ Bx^{2}-6x+1=0, $ the values of A and B such that $ \alpha ,\beta ,\gamma $ and $ \delta $ are in H. P. are
Options:
A) A = 3, B = 8
B) A = -3, B = 8
C) A = 3, B = - 8
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \alpha ,\beta ,\gamma $ and $ \delta $ are in H.P.
$ \Rightarrow ,\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma },\frac{1}{\delta } $ are in A.P.
Let d be the common difference of this A.P.
Now, $ \alpha .\gamma $ are roots of $ Ax^{2}-4x+1=0 $
$ \therefore \frac{\alpha +\gamma }{\alpha \gamma }=\frac{4/A}{1/A}=4 $ or $ \frac{1}{\alpha }+\frac{1}{\gamma }=4 $ i.e. $ \frac{1}{\alpha }+\frac{1}{\alpha }+2d=4 $ or $ \frac{1}{\alpha }+d=2 $ ?. (i) $ \beta ,\delta $ are roots of $ Bx^{2}-6x+1=0 $
$ \therefore \frac{\beta +\delta }{\beta \delta }=\frac{1}{\beta }+\frac{1}{\delta }=\frac{6/B}{1/B}=6 $ or $ \frac{1}{\alpha }+d+\frac{1}{\alpha }+3d=6 $ $ \frac{1}{\alpha }+2d=3 $ ?. (ii) From (i) and (ii), on solving, we get $ \frac{1}{\alpha }=1,d=1.\therefore ,\frac{1}{\alpha }=1,\frac{1}{\beta }=2,\frac{1}{\gamma }=3,\frac{1}{\delta }=4 $ Since, $ \frac{1}{\alpha \gamma }=A\therefore A=3. $ Also, $ \frac{1}{\beta \delta }=B,\therefore B=8 $ Hence $ A=3 $ and $ B=8 $ .
BETA
