JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Analyzing +4 | -1

If positive numbers

{a^{ - 1}},\;{b^{ - 1}},\;{c^{ - 1}}

are in AP, then the product of roots of the equation

{x^2} - kx + 2{b^{101}} - {a^{101}} - {c^{101}} = 0

\left( {k \in R} \right)

View answer and explanation

Correct option: B

< 0

Option B (

<0

) is correct because the product of the roots is equal to

2b^{101} - a^{101} - c^{101}

, which is less than 0 since a, b, c are in HP.

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