Common Difference of arithmetic Progression We are given two arithmetic progression common term of which are $a_n=11+5*(n-1)$ and $b_n=7+3*(n-1)$. What will be the series of the terms formed out of the equal terms of these two progressions.
I have found the equal terms of these series are 16 and 31. So the common difference and the first term of the new series formed out of the two series above should be 15 and 16 respectively. Common difference 15= L.C.M of the common difference of the earlier series. 
But why is the LCM of the previous common differences the new common difference?
 A: You have to solve for which natural numbers $m$,$n$ is $a_m=b_n$. So $11+5(m-1) = 7 + 3(n-1)$. Hence, for which natural numbers $m$, $n$ is $4=5(1-m)+3(n-1) (\star)$. Now take this equation $(\star)$ mod $3$. That gives $1 \equiv 2(1-m) \equiv m-1$ mod $3$, so $m \equiv 2$ mod $3$, say $m=2+3k$. Plugging this back in the $a_m$ gives your new equal terms series $c_k= 11+ 5(2+3k) = 16 + 15k$. And here arises your number $15$. This calculation has to do with the so-called Chinese Remainder Theorem. You might want to take the equation $(\star)$ mod $5$. Check that this of course leads to the same answer.
A: Let $c_n$ denote the series formed by terms common to $a_n$ and $b_n$.  
For any $i < j$, we have $5 \mid (c_j - c_i)$ and $3 \mid (c_j - c_i)$ by virtue of membership in both series.  Thus LCM$(3, 5) \mid (c_j - c_i)$.  So the common difference has to be a multiple of the LCM.
Further, you can note that if $c_1$ is the first common term, $c_1 + k\cdot$LCM$(3, 5)$ will always be in both $a_n$ and $b_n$ for every $k \in \mathbb{N}$., hence the LCM itself is the common difference of $c_n$.
