Finding
the Next Number in a Sequence: Sections: Common differences, Recursions, General examples, Nonmath "sequences"
Sometimes, when dealing with fractions, they've really given you two sequences in one. In this case, looking at the numerators and denominators separately: 1, 3, ___,
7, 9
The numerators appear to be counting by twos; the denominators appear to be doubling. The missing term is ^{5}/_{4}.
This sequence is actually related to the previous one. If I take the sequence terms apart, listing alternate terms in their own sequences, I get: 5, 8, 11, 14, 17,... 2, 3, 4, 5, 6,... In other words, the oddnumbered terms (the first, third, fifth, etc) appear to be of the form "add 3 to the previous term", and the evennumbered terms (the second, fourth, sixth, etc) appear to be of the form "add 1 to the previous term". Then the next two terms are 17 + 3 = 20 and 6 + 1 = 7. The next two terms, in order, are 20 and 7.
If you try common differences on this sequence, you'll see that it doesn't work, so this does not appear to be a polynomial sequence. Whenever a sequence is not obviously generated by a polynomial, it is generally (though not always) generated by a recursion. That is, I should look for some pattern that relates the later terms in this sequence to the earlier ones. At first, I thought the rule for this was going to be "add the preceding terms together", because: 1 + 2 = 3
But 3 + 5 = 8, not 7, so that doesn't work. And multiplication clearly doesn't work, since 2×3 = 6, not 5. However, notice that: Copyright © Elizabeth Stapel 20022011 All Rights Reserved 6 + 5 = 11
I would guess that the formula is meant to be: a_{1}
= 1 There is no guarantee that this is the "right" answer, since the "right" answer is the one that your teacher wants. However, this formula does "work", so it should be acceptable to say that the next term is: The next (nineth) term is 6 + a_{7} = 6 + 13 = 19. Note, however, that the sequence might be simpler even than that. Start listing the first few primes, and see what kind of sequence you get. Compare this result with the values of the above listing.
I checked, and this didn't appear to be a polynomial sequence. It didn't appear at first to be a recursive sequence, but recursives can be messy and hard to figure out. But if one is particularly patient and clever, and plays around with this a bit, one might come up with: a_{1} = 2
The pattern appears to be "take the last term, multiply by three, and then add the number for the sequence term you're on". Continuing, we get: a_{6} = 260×3
+ 6 = 786
So the pattern gives us the seventh term we were supposed to get, which confirms that the sixth term should probably be 786.
For this one, I noticed that 4 is twice 2, and 6 is twice 3. Then 6 (the second one) is thrice 2 and 9 is thrice 3. Continuing, 8 is four times 2, so the next number should be four times 3, or 12. You can also view this as equivalent fractions: 2/3, 4/6, 6/9, 8/12, etc. Either way, the next number is 12. By the way, "fiddling around" is very often the best method for finding solutions. Don't be afraid to grab some scratch paper and just try stuff. There's nothing "wrong" with you if an answer isn't immediately obvious to you. Just keep trying. << Previous Top  1  2  3  4  5  6  7  Return to Index Next >>



Copyright © 2021 Elizabeth Stapel  About  Terms of Use  Linking  Site Licensing 




