if a,b,c are 3 distinct positive numbers in H.P. then the equation x^2 -kx + 2b^101-a^101-c^101=0, k belongs to R

(A) has two distinct real roots.

(B)has product of roots positive

(C)has product of roots negative

(D)has imaginary roots

Hello student,
Please find the answer to your question below
We know that AM>GM
As a,b,c are in HP
(aⁿ+cⁿ)/2>…............(1)
But GM>HM
so>b
or>bⁿ..................(2)
So from 1 and 2(aⁿ+cⁿ)/2>>bⁿ
aⁿ+cⁿ>2bⁿ
S0 a¹⁰¹+c¹⁰¹-2b¹⁰¹>0 so2b^101-a^101-c^101<0...............(3)
So the product of roots have -ve sign
so C is the correct answer