If the roots of the equation `x^(3) - px^(2) + qx - r = 0` are in A.P., then

Options

A. `27r^(2)+9pqr+2q^(3) = 0`
B. `27r^(2) - 9pqr + 2q^(3) = 0`
C. `2r^(2) - 9 pqr + 27 q^(3) = 0`
D. `27r^(2) - 9pqr - 2q^(3) = 0`

Answer

B

Explanation

The equation whose roots are reciprocals of the roots of the given equation is given by
`(1)/(x^(3))-(p)/(x^(2))+(q)/(x)-r=0 rArr rx^(3) - qx^(2) + px - 1 = 0" "...(i)`
Since the roots of the given equation are in H.P. Therefore, the roots of equation (i) are in A.P. Let its roots be a-d, a and a+d. Then,
`(a-d)+a+(a+d)=-(-(q)/(r))rArr 3a = (q)/(r) rArr a = (q)/(3r)`
Since a is a root of equation (i). `therefore" "ra^(3) - qa^(2) + pa - 1 = 0`
`rArr" "r((q)/(3r))^(3)-q((q)/(3r))^(2)+p((q)/(3r))-1 = 0`
`rArr" "(q^(3))/(27r^(2))-(q^(3))/(9r^(2))+(pq)/(3r)-1 = 0`
`rArr" "q^(3) - 3q^(3) + 9pqr - 27r^(2) = 0 rArr 27r^(2) - 9pqr + 2q^(3) = 0`