Multiple choice

Let $ \alpha,\beta,\gamma $ be the roots of the equation ${x}^{3}+{3ax}^{2}+3bx+c=0$. If $ \alpha,\beta,\gamma $are in $H.P$ then $\beta$ is equal to-

  1. $\dfrac{-c}{b}$
  2. $\dfrac{c}{b}$
  3. $-a$
  4. $a$
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A Correct answer
Explanation

If roots are in HP, their reciprocals are in AP. Let roots be 1/p, 1/q, 1/r. The equation for reciprocals is cx^3 + 3bx^2 + 3ax + 1 = 0. For roots in AP, the middle term is related to the coefficients. For a cubic x^3 + Ax^2 + Bx + C = 0, if roots are in AP, the middle root is -A/3. Here, the middle root of the reciprocal equation is -(3b/c)/3 = -b/c. Thus, the middle root of the original equation is -c/b.

AI explanation

Since the roots alpha, beta, and gamma of the cubic equation x^3 + 3ax^2 + 3bx + c = 0 are in harmonic progression, their reciprocals 1/alpha, 1/beta, and 1/gamma must be in arithmetic progression. Using Vieta's formulas, we know alpha + gamma + beta = -3a and alpha gamma + beta gamma + alpha beta = 3b. Because 1/alpha and 1/gamma are symmetric around 1/beta, we have the relation 2/beta = 1/alpha + 1/gamma, which simplifies to 2 alpha gamma = beta (alpha + gamma). Solving for beta using the sum and product relations derived from Vieta's formulas ultimately yields beta = -c/b.

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