If the roots of $x^3-p x^2+q x-r=0$ are in AP. Then,
If the roots of $x^3-p x^2+q x-r=0$ are in AP. Then,
$2 p^3-9 p q+27 r=0$
$2 p^3+9 p q-27 r=0$
$2 p^3-8 p q+27 r=0$
$2 p^3-9 p q+28 r=0$
Solution
Let $a-d, a, a+d$ are roots of
$
x^3-p x^2+q x-r=0
$
Sum of roots $=-\frac{b}{a}$
$
\begin{aligned}
a-d+a+a+d & =-\frac{(-p)}{1} \\
3 a & =p \\
a & =\frac{p}{3}
\end{aligned}
$
Since, $a=\frac{P}{3}$ should be satisfied by given equation.
So, put $x=\frac{p}{3}$ in Eq. (i)
$
\left(\frac{p}{3}\right)^3-P\left(\frac{p}{3}\right)^2+q\left(\frac{p}{3}\right)-r=0
$
$
\begin{aligned}
\frac{p^3}{27}-\frac{p^3}{9}+\frac{p q}{3}-r & =0 \\
p^3-3 p^3+9 p q-27 r & =0 \\
-2 p^3+9 p q-27 r & =0 \\
2 p^3-9 p q+27 r & =0
\end{aligned}
$
Hence, option (1) is correct