Inspired by Hung Woei Neoh
For how many real values of x , does x n , x n + 1 , x n + 2 , ⋯ , follow an arithmetic progression where n is any real value?
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2 solutions
A formal proof to this question:
If you have done the inspiration question first, you would know that for a GP to be an AP at the same time, we must have d = 0 and r = 1 . I will not prove this anymore.
Now, notice that
x n , x n + 1 , x n + 2 , …
is actually a geometric progression where
a = x n and r = x
From the statement above, we know that r must be 1 . Therefore, x = 1
Which means that only 1 value of x satisfies the condition
(P.S. I'm commenting my solution, which means I did it wrongly. I misinterpreted the question XD)
x n , x n + 1 , x n + 2 , . . are in AP
2 x n + 1 = x n + 2 + x n
2 x = x 2 + 1
x 2 − 2 x + 1 = 0
( x − 1 ) 2 = 0
x = 1
∴ x = 1 is the only solution.
Nice solution(+1)
The only value of x for which the arithmetic progression exists is 1 because 1 raised to the power of any real value is 1 and so it follows a constant AP ie. AP woth common difference 0 . Now, what about x = 0 . This AP is broken at 0 0 which is undefined. So, 0 is not an answer. So, there is only 1 value of x for which this arithmetic progression exists.