If one root of `Ax^(3)+Bx^(2)+Cx+D=0,Dne0` is the arithmetic mean of the other two roots, then the relation `2B^(2)+lambdaABC+muA^(2)D=0` holds good. Then, the value of `2lambda+mu` is

Explanation

Given equation, `Ax^(3)+Bx^(2)+Cx+D=0" " ".......(i)"`
where, `Ane 0`
Let roots are `alpha, beta, gamma` then `beta=(alpha+gamma)/(2)" " "…….(ii)"`
Given relation, `2B^(3)+lambdaABC+muA^(2)D=0" " "....(iii)"`
From Eq. (i), `alpha+beta+gamma=-(beta)/(A)`
`implies 3beta=- (B)/(A)" " [" from Eq. (ii) "]`
`implies beta=- (B)/(3A)`
Now, `beta` satisfy Eq. (i), so
`A((-B)/(3A))^(3)+B((-B)/(3A))^(2)+C((-B)/(3A))+D=0`
`implies (-B^(3))/(27A^(2))+(B^(3))/(9A^(2))-(BC)/(3A)+D=0`
`implies (2)/(27)(B^(3))/(A^(2))-(BC)/(3A)+D=0`
`implies 2B^(3)-9ABC+27AD^(2)=0`
Compare with Eq. (iii), we get
`lambda=-9,mu=27`
`2lambda +mu=-18+27=9`.