If the roots of $10 x^{3} - c x^{2} - 54 x - 27 = 0$ are in harmonic progression, then the value of c is

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Solution

The correct option is C 9
$10 x^{3} - c x^{2} - 54 x - 27 = 0$
Let the roots be $α, β, γ . T h e n, α + β + γ = \frac{c}{10}$ ....(1)
$α β + β γ + γ α = - \frac{54}{10} = - \frac{27}{5}$ ...(2)
$α β γ = \frac{27}{10}$ ...(3)
As roots are in H.P., $β = \frac{2 α γ}{α + γ}$ ....(4)
$\Rightarrow α β + β γ = 2 α γ$
Substituting this in (2)
$3 α γ = - \frac{27}{5} \Rightarrow α γ = - \frac{9}{5}$ ....(5)
From (3) $β = - (\frac{27}{10} \times \frac{5}{9}) = - \frac{3}{2}$ ....(6)
Again from (4) $α + γ = - \frac{18}{5} \times - \frac{2}{3} = \frac{12}{5}$ ...(7)
Using (6) and (7) in (1)
$\frac{c}{10} = (α + γ) + β = \frac{12}{5} - \frac{3}{2} = \frac{9}{10} ∴ c = 9$

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