Question

If the roots of $x^3-3x^2+kx+3=0$ are in A.P. then $k=$

• A. $-1$
• B. $1$
• C. $-3$
• D. $3$

Answer 1

Answer: (A)

$-1$

As roots are in A.P then let $a-d,a,a+d$ be the roots of $x^3-3x^2+kx+3=0$
$S_1=3\Rightarrow a=1$
Substituting this in equation, we get
${\left(1\right)}^3-3{\left(1\right)}^2+k+3=0\Rightarrow k=-1$