How to find the meeting number of two sequences? 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: 

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

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