How do I find the same number of following two sequences?

Will these two sequences ever produce any common number? If no, then what will be the possible near number where they are about to meet?

sequence 1:

$1805$, $1814$, $1823$, $\dots$

sequence 2:

$1394$, $1409$, $1424$, $\dots$

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    $\begingroup$ These are actually sequences, not series. $\endgroup$ – Brian M. Scott May 19 '12 at 20:43
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    $\begingroup$ sorry, i am just new to maths, and confused in sequences and series. :( $\endgroup$ – Zain Shaikh May 19 '12 at 20:47
  • $\begingroup$ That’s okay. If it’s just a list, it’s a sequence; if it’s a sum, it’s a series. $1394+1409+1424+1439$, for instance, would be a short arithmetic series. $\endgroup$ – Brian M. Scott May 19 '12 at 20:48
  • $\begingroup$ wow, awesome, just a simple difference it is. thanks :) $\endgroup$ – Zain Shaikh May 19 '12 at 20:56

Hint: The $n$-th term of the first sequence is $1805+9(n-1)$ and the $n$-th term of the second is $1394+15(n-1)$.

  • $\begingroup$ how do i find the nth term where these both sequences meet? $\endgroup$ – Zain Shaikh May 19 '12 at 20:48
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    $\begingroup$ Put my first expression equal to my second expression, solve for $n$. We get $1805+9(n-1)=1394+15(n-1)$, which simplifies to $411=6(n-1)$. This has no solution, since $411$ is not a multiple of $6$. But we get $n-1=411/6=68.5$, giving $n=69.5$, so they are closest at the $69$th term and the $70$th term. A related but very different problem is to describe all the numbers the two sequences have in common, not necessarily at the same place in the sequence. That has been fully dealt with by Brian M. Scott. $\endgroup$ – André Nicolas May 19 '12 at 20:58
  • $\begingroup$ Wonderful. Your solution is excellent and I understood it easily. Thank you so much. :) $\endgroup$ – Zain Shaikh May 19 '12 at 21:04
  • $\begingroup$ @ZainShaikh: The other solution (not mine) solves a different problem than the one that I solved. Please think carefully about whether that might be the problem you wanted the answer to, and if it is, do not hesitate to unaccept my solution and accept the other. $\endgroup$ – André Nicolas May 19 '12 at 21:08
  • $\begingroup$ Thank you, Sir. I tried your solution with different values of a and d and found the perfect value of n. now trying to understand the answer of Sir Brian M. Scott to find the number behind n. $\endgroup$ – Zain Shaikh May 19 '12 at 21:31

Added: This answer is based on the understanding that you want to find the first number that belongs to both sequences, no matter where it occurs in each of the sequences. In other words, if your sequences were $\langle 1,4,7,10,\dots\rangle$ and $\langle 5,7,9,11,\dots\rangle$, my answer would find $7$, even though it’s the third term of the first sequence and only the second term of the second sequence. If you also want the matching number to appear in the same place in both sequences, that’s impossible; see André’s answer for details.

Your first sequence contains the numbers $1805+9k$ for $k=0,1,2,\dots$, and your second contains the numbers $1394+15k$ for $k=0,1,2,\dots$. Thus, the question is whether there are non-negative integers $m$ and $n$ such that $$1805+9m=1394+15n\tag{1}\;.$$ Equation $(1)$ is equivalent to $$15n-9m=411\;,$$ which can be divided by $3$ to yield $$5n-3m=137\;.\tag{2}$$ Equation $(2)$ is a standard linear Diophantine equation; since $3$ and $5$ are relatively prime, it has infinitely many solutions, which may be found with the extended Euclidean algorithm.

Alternatively, if you can find one solution by inspection, you can use Bézout’s identity to write down the general solution. In this case I can see that $5\cdot28-3\cdot1=137$, so one solution is $n=28,m=1$, giving the common term $1394+15\cdot28=1805+9\cdot1=1814$. The general solution is then

$$\left\{\begin{align*} m&=1+5k\\ n&=28+3k \end{align*}\right.\;,$$

and the matching numbers in the two sequences are numbers of the form $1814+45k$ for $k=0,1,2,\dots$. In particular, the first one is the one that I found by inspection.

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    $\begingroup$ Sorry, but I am just beginner and never read about linear Diophantine equation, extended Euclidean algorithm or Bézout’s identity :( $\endgroup$ – Zain Shaikh May 19 '12 at 21:05
  • $\begingroup$ That’s why I included links. The one for Bézout’s identity, in particular, lets you use the identity (as I did above) even if you don’t know why it works or where it comes from. $\endgroup$ – Brian M. Scott May 19 '12 at 21:20
  • $\begingroup$ @Gigili: Fixed; thanks. $\endgroup$ – Brian M. Scott May 19 '12 at 21:30
  • $\begingroup$ Thanks, feeling excited to learning new things :D $\endgroup$ – Zain Shaikh May 19 '12 at 21:30

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