Weird AP
Find the number of real values of such that the three terms , , and form a non-constant arithmetic progression in that order.
The answer is 0.
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1 solution
If we substitute U=2^x,We have that U,U^2,and U^3 is an arithmetic sequence.We will disprove its existence by casework.Case 1:x=0.Then U=1 and the sequence is constant(not allowed in the problem).Case 2 :x is negative.Then the sequence can be multiplied by U^3 to get U^2,U,1.The difference between the first term and the second is U^2-U and the second is U-1.Since this is an arithmetic sequence we know that U^2-U=U-1(note that the differences stay equal despite the multiplication bu U^3)so U(U-1)=U-1.Dividing both sides by U-1 gives us U=1.If 2^x=1,then x must be 0,but we said x must be negative,contradiction.Case 3.x is positive.Using the same difference trick,we have U^3-U^2=U^2-U,so U(U^2-U)=U^2-U,so U=1,x=0,we said x is positive,contradiction. Therefore there is no way for this sequence to exist for any possible value of x.