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How do I mathematically show that the common terms between the series $3+7+11+....$ and $1+6+11+....$ form an arithmetic progression without actually finding all the individual terms. How does the LCM of the common differences of the given series becomes the common difference of the new series?

My Attempt: $$ 3+(n-1)4=1+(m-1)5\implies 4n-1=5m-4\implies5m-4n=3 $$ But this does not tell me the above statement unless I try all the integer combinations of $m$ and $n$.

Sooraj S
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  • hint what is LCM ? –  Sep 01 '17 at 17:39
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    least common multible – Dr. Sonnhard Graubner Sep 01 '17 at 17:42
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    If you don't mind using some heavier machinery, all terms in the sequence $3,7,11,\dots$ satisfy $x\equiv 3\pmod{4}$. Similarly, all numbers in the sequence $1,6,11,\dots$ satisfy $x\equiv 1\pmod{5}$. By the chinese remainder theorem, numbers which satisfy both are then of the form $x\equiv 11\pmod{20}$. It should be clear that the set of numbers satisfying that last condition also form an A.P. As for why the lcm of the differences is the difference of the common series, that follows again from chinese remainder theorem. – JMoravitz Sep 01 '17 at 17:42
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    Why don't you base your calculation on the series starting at the common term you know i.e. 11? – Mark Bennet Sep 01 '17 at 17:43
  • @MarkBennet wht difference can it make ? – Sooraj S Sep 01 '17 at 17:48
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    Well it simplifies the arithmetic a bit because starting at $11$ in both cases (or any common term) you simply need to satisfy $4m=5n$ and that is easy to do. $11$ will be part of your target progression and you aren't actually concerned at all about how the terms in the original progressions are numbered (which term is "first") - just identifying the common terms. – Mark Bennet Sep 01 '17 at 17:51
  • ohh thts right. I think LCM part makes sense now. thnx. – Sooraj S Sep 01 '17 at 17:58

3 Answers3

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Continuing from your work you can see that the final equation is a simple linear Diphantene equation. Solving it for $m$ (you can also solve it with respect to $n$) you will get $m = 4t + 3$. Plugging it into the coresponding form of the sequence we get that the common terms are given by:

$$1 + 5(m-1) = 1 + 5(4t + 3 - 1) = 20t + 11$$

Hence the common terms make an arithemtic progression and it's given by $c_n = 20n + 11$.

Stefan4024
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  • thnx, i think this is what i was looking for. but how do you write $m=4t+3$ from the above equation. Could u pls help me understand. – Sooraj S Sep 01 '17 at 18:20
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    As I've said in my answer what you got is a linear Diphantene Equation and you can easily find basic tutorials on how to solve them on the Internet. Anyway in short we have $5 - 4 = 1 \iff 5 \times 3 - 4 \times 3 = 3 \iff 5(4t + 3) - 4(5t + 3) = 3$. So this will give you the form of $m$ and $n$. – Stefan4024 Sep 01 '17 at 18:27
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The reason LCM is the difference, is suppose we have 1 value they intersect at already, to be in the first sequence they have to have a difference that's a multiple of the first sequences difference. For the second sequence it's the same deal, to be in that sequence it must have a difference that divides by the second sequences difference and LCM of the differences is the least difference that fits both of these).

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This is as far as I go: $$a_{n}=\sum ^{n}_{k=1}(4k-1)=\dfrac{3+(4n-1)}{2}n=2n^{2}+n$$ $$b_{m}=\sum^{m}_{k=1}(5k-4)=\dfrac{1+(5m-4)}{2}m=\dfrac{5}{2}m^{2}-\dfrac{3}{2}m$$ $$a_{n}=b_{m}\rightarrow 2n^{2}+n=\dfrac{5}{2}m^{2}-\dfrac{3}{2}m\rightarrow 4n^{2}+2n=5m^{2}-3m$$ Who ensures that common terms are in progression?

Guillemus Callelus
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