Let us say I have two Arithmetic progressions

$AP_1 = a + nb$

$AP_2 = c + nd$

Can we find a new Arithmetic progression $AP_3$ which is intersection of both $AP_1$ and $AP_2$?

  • $\begingroup$ If there is one common term, yes. But they could be disjoint - e.g. Odd numbers and even numbers. $\endgroup$ – Macavity Jun 4 '15 at 16:21
  • $\begingroup$ Not always. But for example if $b$ and $d$ are relatively prime (no common divisor greater than $1$), we can. $\endgroup$ – André Nicolas Jun 4 '15 at 16:21
  • $\begingroup$ if $AP_1 = a + nb$ where $a=2,b=3$ then we have $AP_1 = 2+3=5,2+6=8,2 + 9 =11 .....$ Now if $AP_2 = c + nd$ where $c=4,d=5$ then we have $AP_2 = 4 + 5=9,4+10=14,4+15$ What I am saying is that the intersection can be empty !! $\endgroup$ – alkabary Jun 4 '15 at 16:23
  • $\begingroup$ @alkabary Wrong e.g. for that conclusion. $14, 29, ...15k-1,...$ is the intersection, another AP. In effect you are solving $a = \pmod b, c=\pmod d$ simultaneously, which is always solved if $\gcd(b,d) \mid (a-c)$. $\endgroup$ – Macavity Jun 4 '15 at 17:05
  • $\begingroup$ I see that there can be an intersection, but the intersection can not be an Arithmetic progression, if so can we find an equation, for the new series. $\endgroup$ – Naks Jun 4 '15 at 17:16

We have that for any two arithmetic progressions, $AP_1, AP_2$, we have that their intersection are those points where

$b(n_1)+a = d(n_2)+c \iff b(n_1)= d(n_2)+c-a $

for some $n_1, n_2 \in \mathbb{Z}$. If no such $n$s exist, then the intersection is empty. If they exist, we can proceed as follows.

This means that the intersection of these two sets produce the set $AP_3 =(b\cdot d)(n) + (d \cdot k + c)$ for some $k \in \mathbb{Z}$ such that $b \cdot k_1 = (d\cdot k + (c-a))$ for some $k_1 \in \mathbb{Z}$.

We can see this: $(b\cdot d)\cdot(n) + (d \cdot k + c) = (b)\cdot (d\cdot n) + (b \cdot k_1 + a) = (b)\cdot (d\cdot n + k_1) + a$, which is clearly a subset of $AP_1$.

We also see that $(b\cdot d)\cdot(n) + (d \cdot k + c) = (d)\cdot (b\cdot n + k) + c$, which is clearly a subset of $AP_2$, as required.

  • $\begingroup$ Is it necessary for both ap to have same number of terms for forming a new ap $\endgroup$ – Jack Rod Oct 12 '19 at 5:50

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