Multiple choice

Condition for the roots of the equation $x^{3}-px^{2}+qx-r=0$ are in $\mathrm{G}.\mathrm{P}$ is

  1. $q=pr$
  2. $q^{2}=p^{2}r$
  3. $q^{3}=pr^{3}$
  4. $q^{3}=p^{3}r$
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D Correct answer
Explanation

If roots a/r, a, ar are in G.P., their product is a^3 = r, so a = r^(1/3). The sum of roots is a(1/r + 1 + r) = p. The sum of roots taken two at a time is a^2(1/r + 1 + r) = q. Dividing the second by the first gives a = q/p. Substituting a^3 = r gives (q/p)^3 = r, which simplifies to q^3 = p^3 * r.

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