Find of two arithmetic progressions meet

Question

I was working a problem to find if two AP meet or not and came across the below post Formula to find if two AP meet

and in that the formula given by user is

$$A_n=A_1+(n−1)d$$

$$B_m=B_1+(m−1)D$$

and if $A_n=B_m$ then they meet. But somehow it failed for my case of $A_1=14,d=4 B_1=98,D=2$. Could you please tell me where i went wrong or the formula was wrong?

Answer 1

The $2$ sequences meet, but with $n$ being $22$, not $2$, for $m$ being $1$, with them meeting at $98$, as can be easily confirmed by plugging the values into the equations: $$14 + \left(22 - 1\right)4 = 14 + \left(21\right)4 = 14 + 84 = 98$$ $$98 + \left(1 - 1\right)2 = 98 + 0 = 98$$ Note to determine $n$ and $m$, subtracting the second general formula from the first general one provided in the question gives: $$84 = \left(n - 1\right)4 - \left(m - 1\right)2$$ $$84 = 4n - 4 - 2m + 2$$ $$86 = 4n - 2m$$ $$43 = 2n - m$$ Since both $m$ and $n$ must be $\geq 1$, so this limits what there values can be. As $43$ is odd and $2n$ is even, the values of $m$ must all be odd. The smallest odd value is $1$, with this giving $2n = 43 + 1 = 44$, so $n = 22$.

For each increase in $n$, the right side increases by $2$, so $m$ must increase by $2$ (as it is being subtracted) to compensate to keep it an equation. Thus, the sets of $\left(n,m\right)$ values which work are $\left(22,1\right),\left(23,3\right),\left(24,5\right),\ldots$, i..e, in general being $(22 + a, 1 + 2a)$ for $a$ being non-negative integers. This keeps going up for $a = 16$ giving $\left(38,33\right)$, resulting in $162$ as stated by the question's author, hardyz009, in his/her comment to the original question.

Answer 2

There are too many variables involved to make that solution useful. Instead of looking at his results, I think it would be better to apply his method of solution to your particular problem.

\begin{align} 14 + 4(m-1) &= 98 + 2(n-1) \\ 4m + 10 &= 2n + 96 \\ 2m + 5 &= n + 48 \\ 2m - n &= 43 \end{align}

In order for there to be a solution, we must have $\gcd(2, -1) \mid 43$. Which is true.

A solution would be $(m,n)=(22,1)$.

So, the general solution is therefore $(m,n)=(22+t, 1+2t)$ for all integers $t \ge 0$.

We compute

$$14+4(22+t-1) = 98 + 4t$$ $$98 + 2(1+2t-1) = 98 + 4t$$

Answer 3

Using the parameters provided, the expressions for $A_n$ and $B_m$ may be written as $$A_n=4n+10=14, 18, 22, 26, \cdots\;\;\;\\ B_m=2m+96=98, 100, 102, 104, \cdots$$ hence the two series will "meet" at many points, e.g., $98, 102, 106, \cdots$ .

If "meeting" is taken to mean that $m=n$, i.e. where $A_n=B_n$, then this can be solved as two simultaneous equations,$y=4x+10$ and $y=2x+96$, giving $x=43$, hence the solution is $$A_n=B_n=182$$ where $n=43$.