Question:
If zeros of the polynomial $f(x)=x^{3}-3 p x^{2}+q x-r$ are in A.P., then
(a) $2 p^{3}=p q-r$
(b) $2 p^{3}=p q+r$
(c) $p^{3}=p q-r$
(d) None of these
Solution:
Let $a-d, a, a+d$ be the zeros of the polynomial $f(x)=x^{3}-3 p x^{2}+q x-r$ then
Sum of zeros $=\frac{-\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$
$(a-d)+a+(a+d)=\frac{-(-3 p)}{1}$
$3 a=3 p$
$a=\frac{3}{3} p$
$a=p$
Since $a$ is a zero of the polynomial $f(x)$
Therefore,
$f(a)=0$
$a^{3}-3 p a^{2}+q a-r=0$
Substituting $a=p$. we get
$p^{3}-3 p(p)^{2}+q \times p-r=0$
$p^{3}-3 p^{3}+q p-r=0$
$-2 p^{3}+q p-r=0$
$q p-r=2 p^{3}$
Hence, the correct choice is $(a)$