Son asked such a task. Continue the number sequence
5, 7, 8, 12, 11....
The son is in the second grade. They studied the operations of addition, subtraction and multiplication
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2 Answers
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Seems like a pretty bad question to ask anyone you are trying to teach mathematics, but a best guess would be:
$(5 + 3\times 0 = 5), $
$(7 + 5\times 0 = 7), $
$(5 + 3\times 1 = 8), $
$(7 + 5\times 1 = 12), $
$(5 + 3\times 2 = 11), $
$(7 + 5\times 2 = 17), $
$...$
If we index the sequence, say $a(n)$, starting at $a(0) = 5$ we can say
Let: $m = n\mod 2$
($m$ is $1$ if $n$ is odd, and $0$ if $n$ is even)
$a(n) = m\times 7 + (1-m)\times 5 + (m\times 5 + (1-m)\times 3) \times \lfloor\frac{n}{2}\rfloor$
I wonder if they were taught about "flooring" or modulo arithmetic too?
answered Sep 15, 2017 at 17:26
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1$\begingroup$ This is the same as...if we consider alternating sequences and find the next term. Here it would be 5,8,11 form the first with a diff. of 3 and the next one (7,12,__) with a diff of 5 thereby giving 17 as the next term $\endgroup$ Commented Sep 15, 2017 at 17:39
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$\begingroup$ Maybe that is how they are meant to crack it, but it's very weak I think. $\endgroup$ Commented Sep 15, 2017 at 17:41
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1$\begingroup$ Well I doubt it is ...5, 7, 8, 12, 11, 9, 13, 15, 22, 21, ... - that is this :) $\endgroup$ Commented Sep 15, 2017 at 17:44
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Based on the original sequence
5, 7, 8, 12, 11, ..
it seems like the odd-numbered values increase by three, and the even-numbered values increase by five. So I would extrapolate that the series would continue like this:
5, 7, 8, 12, 11, 17, 14, 22, 17, 27, 20 ..
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$\begingroup$ ^vote with a note: I first took "odd-numbered values" to mean 5, 7 and 11 but then realized that it meant odd-position values. $\endgroup$– humnCommented Sep 15, 2017 at 20:46
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$\begingroup$ I do not like this answer, as it breaks the sequence into two unique sequences, that do not rely on each other. $\endgroup$– Jason VCommented Sep 15, 2017 at 21:43
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2$\begingroup$ This is exactly equivalent to Jonathan Allan’s answer (which was posted two hours earlier) but with a slightly different presentation. $\endgroup$ Commented Sep 16, 2017 at 2:27