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Question

If a,b,c,d are four distinct real numbers which are in an A.P., then the smallest positive value of k satisfying 2(ab)+k(bc)2+(ca)3=2(ad)+(bd)2+(cd)3 is

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Solution

Given : a,b,c,d are in an A.P.
Assuming the common difference be D,
Now,
2(ab)+k(bc)2+(ca)3=2(ad)+(bd)2+(cd)32(D)+k(D)2+(2D)3=2(3D)+(2D)2+(D)34D+(k4)D2+9D3=0D(9D2+(k4)D+4)=0
As a,b,c,d are distinct, so D0
9D2+(k4)D+4=0
As DR
(k4)24(4)(9)0k28k1280(k16)(k+8)0k(,8][16,)

Hence, the smallest positive value of k is 16.

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