The roots of equation $15 x^{3} + c x^{2} + 36 x + 8 = 0$ are in H.P., then the value of $c$ is

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Solution

$15 x^{3} + c x^{2} + 36 x + 8 = 0$
replacing $x$ by $\frac{1}{x}$ , we get
$\frac{15}{x^{3}} + \frac{c}{x^{2}} + \frac{36}{x} + 8 = 0$
$\Rightarrow 8 x^{3} + 36 x^{2} + c x + 15 = 0 (1)$
Roots of equation $(1)$ are in A.P.
Let roots be $α - β, α, α + β$ then
$α - β + α + α + β = - \frac{36}{8} = - \frac{9}{2}$
$\Rightarrow 3 α = - \frac{9}{2} \Rightarrow α = - \frac{3}{2}$
This satisfies the equation $(1)$
So,
$8 \times - \frac{27}{8} + 36 \times \frac{9}{4} - \frac{3 c}{2} + 15 = 0$
$\Rightarrow - 27 + 81 - \frac{3 c}{2} + 15 = 0$
$\Rightarrow \frac{3 c}{2} = 69 \Rightarrow c = 46$

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