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I got a question during an interview in a company and I still could not find out the solution. Kindly help me to solve this.

H,A,H,L,U,?

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    $\begingroup$ You mean "letter", not "alphabet". An alphabet is a set of letters. Not the first time I see this mistake. What translator did you use? May be worth sending some feedback. $\endgroup$ – Arnaud Mortier Mar 21 at 16:02
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    $\begingroup$ Was the question asked in English? Because these letters could be the first letters of a common set in another language eg( M,T,W,T,F, ?) would be S for Saturday. $\endgroup$ – Chris Cudmore Mar 21 at 16:20
  • $\begingroup$ 8,1,8,12,21,? positions of the above letters in alphabet. i need to find the next position in the series $\endgroup$ – Arunkumar Thangavel Mar 21 at 16:41
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    $\begingroup$ what was the job? it might be relevant :) $\endgroup$ – JMP Mar 21 at 17:26
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    $\begingroup$ @ArnaudMortier I think it's not so much a mistake as a dialect difference: in Indian English, "alphabet" means what "letter" means in most varieties of English. $\endgroup$ – Gareth McCaughan Mar 21 at 18:17
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I believe that the answer is

L

This is based on the fact that the first and last letters of each triplet is the same. For example, HAH, LUL, etc...

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This is my random solution

H,A,H,L,U,___

The answer is:

W

 

Sequence is 8, 1, 8, 12, 21 - ignoring the first as red herring, the distance between each value is [7,4,9] (for example, 8 - 1 = 7, 12 - 8 = 4, etc.). The distance between those values shows a growing sequence of odd, prime numbers: [3,5] (for example, 7-4 = 3, 9-4 = 5).

 

However, it seems it oscillates between subtraction and addition. 7 - 4 = 3, but 4 + 5 = 9. I'd assume the next sequence would be 7, which is an odd prime number. And the next step requires subtraction, so it is: 9 - 7 = 2. Therefore, the next number would jump by 2 units (21 + 2). Which ends up in 23, which is the letter W

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    $\begingroup$ If you allow for such complicatedness in the answer, starting from such a small data set, then there are hundreds of similar valid answers. $\endgroup$ – Arnaud Mortier Mar 22 at 12:55

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