Multiple choice

The sum of the roots of quadratic equation $ax^2 + bx + c = 0 \,(a, b, \neq 0)$ is equal to the sum of squares of their reciprocals, then $\dfrac {a}{c}, \dfrac{b}{a}$ and $ \dfrac {c}{b}$ are in

  1. AP

  2. GP

  3. HP

  4. None of these
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C Correct answer
Explanation

Let the roots be alpha and beta. Using Vieta's relations and the given condition involving the reciprocals, the reciprocals of a/c, b/a, and c/b form an arithmetic progression. Therefore, the original three quantities are in harmonic progression.

AI explanation

Using Vieta's formulas for ax^2 + bx + c = 0, the sum of the roots is -b/a and the product of the roots is c/a. The problem states that the sum of the roots equals the sum of the squares of their reciprocals, giving -b/a = (1/r1 + 1/r2)^2 - 2/(r1*r2). Substituting the sum and product gives -b/a = (a^2/c^2) - 2a/c. Multiplying by -a/c, we get b/c/a = -a/c + 2. This rearranges to 2 = a/c + b/a. Dividing by 2 yields 1 = a/2c + b/2a, showing that a/2c and b/2a are symmetric around 1. Since a/c, b/a, and c/b are the reciprocals of c/b, a/b, and b/c, they must be in Harmonic Progression.

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