Progressions: A Unification!
It is given that there exists a set of real numbers , such that its elements taken in some order form a non-constant arithmetic progression and its elements taken in some other order form a non-constant geometric progression .
What is the maximum possible cardinality of this set ?
If you think that there is no upper bound for its cardinality, input -1 as your answer.
If you are looking for more such twisted questions, Twisted problems for JEE aspirants is for you!
The answer is 3.
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1 solution
Any 2 numbers are always in AP or GP.
If Ξ only contains positive numbers then the maximum numbers which can satisfy is 2 (can be proved using AM GM and then using the fact that it is a nonconstant progression)
So , If Ξ contains both positive and negative numbers, arranging them in ascending order will give me the AP. Also changing all the signs will create a new set which will satisfy both conditions.
Let 1st term of GP be a and its ratio be − r , r > 1
We have the terms as a , − a r , a r 2 . . .
Case 1: 3 terms
AP will be − a r , a , a r 2 So 2 a = a r 2 − a r
Thus r 2 − r − 2 = 0
And ( r − 2 ) ( r + 1 ) = 0
We can see that r = − 1 gives constant progression (and r > 1 ) So only solution is r = 2 .
Case 2: more than 3 terms.
If we have more than 3 terms then you can see that if we arrange the terms again in the order for AP, We get that Again the 1st 3 terms of GP must be an AP. So the common ratio must be 2. But for r = 2 the 4th term does not correspond to forming an AP (4th term is − a r 3 = − 8 a )(-8a,-2a,a,4a do not form an AP) If 4th term does not form the AP then for no n > 3 will it form an AP.