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

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