If `1/a,1/b,1/c` are in A.P and a,b -2c, are in G.P where a,b,c are non-zero then

Answer

A, B, C, D

Explanation

we have `b=(2ac)/(a+c)` and `b^(2)=-2ac`
`thereforeb=(-b^(2))/(a+c)`
`rArr a+b+c=0`
`rArr2b-2a-2c`
Therefore, -2a,b,-2c are in A.P.
Also, `a^(3)+b^(3)+c^(3)=3abc`
Since `b^(2)=-2ac,b^(4)=4a^(2)c^(2)`
So, `a^(2),b^(2),4c^(2)` are in G.P.
Since a+b+c=0, equations `ax^(2)+bx+c=0` has one root x=1
So, equation has real roots.