If the roots of the equation 10x3−cx2−54x−27=0  are in H.P., then c=

The given equation is 10x3−cx2−54x−27=0 ….. ( 1 ) Given that, roots of Eq. ( 1 )  are H.P. Then roots of 27x3+54x2+cx−10=0  are in A.P.   ….. ( 2 )Let the roots to be α−β, α, α+β  Then sum of the roots:  α−β+α+α+β=−5427 ⇒α=−23 Substitute α=−23  in Eq. ( 2 ), we get⇒ 27(−827)+54(49)−2c3−10=0 ⇒ −8+24−2c3−10=0 ⇒c=9