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Question

Let three positive real numbers a,b,c are in A.P. If abc=4 and the minimum value of b is 2k, then the value of 6k is 


Solution

Let the common difference be d
b=a+da=bdc=b+d
Now, abc=4
(bd)b(b+d)=4b(b2d2)=4
As b2b2d2,
b×b2b×(b2d2)b34b22/3
Therefore, the minimum of b is 22/3.
Hence, 6k=4

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