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 $n$.
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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?
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 $c_n = 20n + 11$.