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Let `a ,b ,c ,d` be four distinct real numbers in A.P. Then half of the smallest positive valueof `k` satisfying `a(a-b)+k(b-c)^2=(c-a)^3=2(a-x)+(b-d)^2+(c-d)^3` is __________.

Text Solution

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The correct Answer is:
16
Since a,b,c,d are in A.P., we have
b-a=c-b=d-c=D (let common difference)
or d=a+3D
`rArra-d=-3D and d=b+2D`
or b-d=-2D
Also c=a+2D or 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+kD^(2)+(2D)^(3)=-6D+4D^(2)-D^(3)`
`rArr9D^(2)+(k-4)D+4=0`
Since D is real, we have `(k-4)^(2)-4(4)(9)ge0`
or `k^(2)-8k-128ge0` or `(k-16)(k+8)ge0`
`thereforekin(-oo,-8]cup[16,oo)`
Hence, the smallest positive value of k = 16.
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