Question

Let $a,b,c,d$ be four distinct real numbers in A.P. Find the smallest positive value of $k$ satisfying $2(a-b)+k(b-c{)}^2+(c-a{)}^3=2(a-d)+(b-d{)}^2+(c-d{)}^3$.

Answer 1

Answer: 0016

Since $a,b,c,d$ are in A.P.
$\therefore b-a=c-b=d-c=D$ (let common difference)
$\Rightarrow d=a+3D\Rightarrow a-d=-3D\text{ and }d=b+2D\Rightarrow b-d=-2d$
Also $c=a+2D\Rightarrow c-a=2D$
$\therefore$ Given equation $2(a-b)+k(b-c{)}^2+(c-a{)}^3=2(a-d)+(b-d{)}^2+(c-d{)}^3$ becomes $-2D+k{D}^2+(2D{)}^3=-6D+4D^2-D^3$
$\Rightarrow 9D^2+(k-4)D+4=0$
Since D is real $\Rightarrow (k-4{)}^2-4(4)(9)\ge 0$
$\Rightarrow k^2-8k-128\ge 0\Rightarrow (k-16)(k+8)\ge 0$
$\therefore k\in (-\infty ,-8]\cup [16,\infty )$
Hence smallest positive value of $k=16$