I want to share a question that is created by myself.

I will give a hint in 24 hours and my answer in 3 days given that nobody could answer my question.

Here is my number sequence:

$3/2, 2/3, 3/4, 1/3, 3/8, ?$

If you guys want some extremely challenging questions, please check these two questions posted by me:

Number sequences: 6X000X9, 700XX08,00000015,?,00000015

What are the alphabets in the question mark?

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    $\begingroup$ There is only 1 answer and the logic must be the simplest and neatest. $\endgroup$ – LETTERKING May 16 '19 at 15:57

Obviously the answer is



The even and odd terms are multiplied by $ 1/2 $.

That is,

For odd:
$3/2 \times 1/2 = 3/4.$
$3/4 \times 1/2 = 3/8.$

For even:
$2/3 \times 1/2 = 2/6\implies 1/3.$
$1/3 \times 1/2 = 1/6.$

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I think the answer is

$\frac{2}{3}$ or $\frac{1}{6}$


For answer 1:
Each even term is the reciprocal of the preceding odd term. If the fraction is still reducible, then the remainder is the even term.
$2/3$ is irreducible, so it is the second term. $4/3 = 1 + 1/3$. Thus, $1/3$ is the fourth term.
Similarly, $8/3 = 2 + 2/3$. So, the sixth term should be $2/3$ $$$$
For answer 2:
$A_n = \frac{1}{2}A_{n-2}$ [a simpler version posted by @Peregrine Rook. Thanks]

So we have \begin{array}lA_1 = 3/2\\A_2 = 2/3\\A_3 = \frac12A_1 = \frac12\times 3/2=3/4\\A_4 = \frac12A_2 = \frac12\times 2/3=1/3\\A_5 = \frac12A_3 = \frac12\times 3/4=3/8\\A_6 = \frac12A_4 = \frac12\times 1/3=1/6 \text{(the answer)}\phantom{WWWWWWWWWWWWWWWW}\\ ~~\vdots\end{array}

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  • 2
    $\begingroup$ One of your answers is correct! But the logic is wrong sorry. $\endgroup$ – LETTERKING May 16 '19 at 15:56
  • $\begingroup$ @PeregrineRook Thanks for your comment. But at that time I didn't get this one. $\endgroup$ – Ak19 Jun 3 '19 at 6:47
  • $\begingroup$ @PeregrineRook I got it now. Thanks! $\endgroup$ – Ak19 Jun 3 '19 at 7:04
  • $\begingroup$ I believe that it’s clearer now.   If you don’t like it, you are of course free to roll it back. $\endgroup$ – Peregrine Rook Jun 3 '19 at 7:37
  • $\begingroup$ Thanks I'm fine with it ! $\endgroup$ – Ak19 Jun 3 '19 at 7:41

The answer is


For this kind of question, we cannot just simply split the terms as every term is connected to each other.

Therefore, the logic is


Here is the process:






Hope this clarifies.

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  • 1
    $\begingroup$ (It is generally nicer to give a solver some hints and some time to let them come up with the intended full solution themselves; if their answer is substantially close, you might instead annotate their final solution with, e.g., "Added by OP" notes that fill in missing or incorrect details. Having said that, in this case it's rather hard to argue how the answer offered by @Stojan Samojlovski is in any way lacking; it has the intended answer via a well justified mechanism that is ultimately equivalent to the method your own answer describes.) $\endgroup$ – Rubio May 17 '19 at 9:58
  • $\begingroup$ Okay, I get it! Next time I will give a hint first. I am still new here sorry... $\endgroup$ – LETTERKING May 17 '19 at 10:27
  • 1
    $\begingroup$ (1) You even said that you would wait three days before posting your answer.  (2) Stojan Samojlovski got the right answer, with logic that’s mathematically equivalent to yours, and your explanation isn’t really much simpler or neater.  You should have just accepted their answer.   (For that matter, Ak19 pretty much got it right, also, but their explanation is very unclear.) $\endgroup$ – Peregrine Rook Jun 3 '19 at 6:44

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