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If the roots of 10 x^3-n x^2-54 x-27=...

If the roots of `10 x^3-n x^2-54 x-27=0` are in harmonic oprogresion, then `n` eqauls _________.

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The correct Answer is:
9

`10x^(3)-nx^(2)-54x-27`=0 has roots in H.P.
put x=1/t
`27t^(3)+54t^(2)+nt-10=0`
This equation has roots in A.P. Let the roots be a-d,a, and a+d
`therefore3a=-(54)/(27)` or `a=-2/3`
Also `(a-d)a(a+d)=10/27`
`therefore2/3(4/9-d^(2))=-10/27` or `(4/9-d^(2))=-5/9`
`therefored^(2)=1` or `d=pm1`
So, roots are `1/3,-2/3,-5/3`
`thereforen/27=10/9-5/9-2/9` or `n/27=3/9`
or n=9
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