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The condition that ensures that the roots of the equation $x^3-px^2+qx-r=0$ are in $H.P.$ is

  1. $r^2-9pqr+q^3=0$
  2. $27r^2-9pqr+3q^3=0$
  3. $3r^3-27pqr-9q^3=0$
  4. $27r^2-9pqr+2q^3=0$
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Lets  $\alpha,\beta$ and $\gamma$ are the roots of $ax^{3}+bx^{2}+cx+d=0,$ then

$\alpha+\beta+\gamma = \dfrac{-b}{a}$

$\alpha\beta+\beta\gamma+\alpha\gamma = \dfrac{c}{a}$

$\alpha\beta\gamma = \dfrac{-d}{a}$

_____________________________________________________________

Lets  say roots of the equation $x^{3}-px^{2}+qx-r = 0 $  are $\alpha,\beta$ and $\gamma$ are  in $H.P.$

So, $\dfrac{1}{\alpha},\dfrac{1}{\beta},\dfrac{1}{\gamma}$ are in $A.P.$

We know that if $A,B,C$ are in $A.P.$ then $2B=A+C$

$\implies\dfrac{2}{\beta}=\dfrac{1}{\alpha}+\dfrac{1}{\gamma}$

$\implies\dfrac{2}{\beta}=\dfrac{\gamma+\alpha}{\alpha\gamma}$

$\implies 2\alpha\gamma = \beta\gamma+\alpha \beta\rightarrow (1)$

Now,  

$\alpha+\beta+\gamma = \dfrac{-b}{a} = \dfrac{-(-p)}{1}=p\rightarrow(2)$

$\alpha\beta+\beta\gamma+\alpha\gamma = \dfrac{c}{a}=\dfrac{q}{1}=q\rightarrow(3)$

$\alpha\beta\gamma = \dfrac{-d}{a}=\dfrac{-(-r)}{1}=r\rightarrow(4)$

From the equation $(1)$ and $(3),$ we get

$\underbrace{\alpha\beta+\beta\gamma}+\alpha\gamma =q$

$\implies 2\alpha\gamma + \alpha\gamma=q$

$\implies 3\alpha\gamma =q$

$\implies \alpha\gamma =\dfrac{q}{3}\rightarrow(5)$

From the equation $(4)$ and $(5),$ we get

$\alpha\beta\gamma = r $

$\implies\underbrace{\alpha\gamma}\beta = r $

$\implies\dfrac{q}{3}\beta = r $

$\implies\beta=\dfrac{3r}{q}$

From the equation $(2),$ we get

$\alpha+\beta +\gamma = p$

$\implies \alpha +\gamma = \dfrac{pq-3r}{q}\rightarrow(6)$

From the equation $(1),$ we get

$2\alpha\gamma = \beta\gamma+\alpha \beta$

$\implies \underbrace{2\alpha\gamma} =\underbrace{ \beta}\left[\gamma+\alpha\right]$

$\implies \dfrac{2q}{3}=\dfrac{3r}{q}\left[ \alpha+\gamma\right]$

$\implies \alpha + \gamma=\dfrac{2q^{2}}{9r}\rightarrow (7)$

Now, from the equation $(6)$ and $(7),$ we get

$\alpha +\gamma = \dfrac{pq-3r}{q}=\dfrac{2q^{2}}{9r}$

$\implies \dfrac{pq-3r}{q}=\dfrac{2q^{2}}{9r}$

$\implies 9pqr-27r^{2}=2q^{3}$

$\implies 27r^{2}-9pqr+2q^{3}=0$

So, the correct answer is $(D)$.
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