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I'm looking at three sequences from the test at http://free.ultimaiq.net/nse.htm. I managed to answer 27 of the 30 questions, but I'm really curious about the other three.

Q6: 1001, 4524, 4299, 3984, ? -- here is something like 100 145 244 299 398 45+99+55+99 and the next one should be 398+65and+99 but its not

Q24: 50625, 1296, 8100, 3240, ? -- I only know that the answer is not an integer and there is a hint "think how you can arrive at an+2 given an+1 and an".

Q22: 38, 50, 22, 18, ?, ?

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  • $\begingroup$ Q22 has the property that if you count the 7-segment LED bits set, you get 12,11,10,9, but I can't see any reason behind the choices. $\endgroup$ – JMP Nov 8 '19 at 18:40
  • $\begingroup$ Yes i try many combination's to solve this question and still but thank you for that 2Q. $\endgroup$ – user63706 Nov 8 '19 at 18:44
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Q6:

The answer is 4354, which is +45,+99,+55,+99,+45,+99, so 443,542 splits into the 4-digit 4354.

Q24:

$50625=2^03^45^4$
$1296=2^43^45^0$
$8100=2^23^45^2$
$3240=2^33^45^1$

The power of $3$ stays constant, the powers of $2$ and $5$ are the average of the previous two, so the answer is $2^{2.5}3^45^{1.5}=5122.889809472774517838207541981$, although anything with 1dp or more seems to work.

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  • $\begingroup$ This is pretty stupid. (I don't mean that your answer is wrong, I mean the question is stupid.) Quite aside from the gratuitous "frame shifting", this boils down to asking the solver to extrapolate the sequence 45, 55, ? which could perfectly well yield 65 (OP's guess), 66 (triangular numbers), 45 (your answer), and any number of other roughly equally plausible possibilities. [EDITED to add:] I'm referring to Q6. Q24 is more reasonable. $\endgroup$ – Gareth McCaughan Nov 8 '19 at 18:06
  • $\begingroup$ Thank you JMP for help. $\endgroup$ – user63706 Nov 8 '19 at 18:26
  • $\begingroup$ @JMP The much simpler way to explain the solution for Q24 is that a_n = sqrt(a_n-1 * a_n-2) (i.e. the geometric mean). $\endgroup$ – my pronoun is monicareinstate Nov 9 '19 at 13:18

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