1
$\begingroup$

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 $AP_2$?

$\endgroup$
6
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.