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Question

If a1,a2,a3,a4,a5 are the roots of the equation 6x541x4+97x397x2+41x6=0, such that |a1||a2||a3||a4||a5|, then which of the following is/are correct?

A
a1,a2,a3 are in G.P.
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B
a1,a2,a3 are in H.P.
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C
a3,a4,a5 are in A.P.
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D
the equation has three real roots and two imaginary roots.
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Solution

The correct option is C a3,a4,a5 are in A.P.
This is a reciprocal equation of odd degree with opposite signs.
x=1 is a root.
Dividing L.H.S. by x1, the given equation reduces to
6x435x3+62x235x+6=0
Divide above equation by x2,
6(x2+1x2)35(x+1x)+62=0
Put x+1x=y
(x2+1x2)=y22
6(y22)35y+62=0(3y10)(2y5)=0y=x+1x=103 or 52
3x210x+3=0 or 2x25x+2=0(3x1)(x3)=0 or (2x1)(x2)=0x=13,3 or x=12,2
(a1,a2,a3,a4,a5)=(13,12,1,2,3)
a3,a4,a5 are in A.P.
a1,a2,a3 are in H.P.

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