Question is as follows:
Find formula ($a_{n}=a_1+d(n-1)$) for shared terms of $8,11,14,…$ and $2,7,12,…$ ?
My Work :
I found $a_1$ by continuing sequences and then found $a_2$ which gave me $a_2-a_1=d$ and found the answer.
Problem:
Looked up the answer and saw something I couldn't understand (final formula was same though).
Book's approach:
$d $ is $d = 3\times5$ or, in other words, is LCM of common differences ($ LCM(d_1,d_2)$). And after that found $a_1$ by extending sequence and done.
Confusion :
That should be okay when we have no constant added (like when it's $3x$ or $5x$) because its just like LCM of $3$ and $5$ but I can't understand why it works when there is $3x+5$ and $5x-3$.
Search result:
I get Shared terms in an arithmetic sequence hint, and I've done this as my second option. Yet it's not always fast enough to find those numbers because of gaps that can get bigger than few numbers.
Best Answer
Equate the common terms: $$8+3n=2+5m \Rightarrow 5m-3n=6 \Rightarrow m=3+3k,n=3+5k$$ When $k=0$ (note $m,n\ge 0$) we get: $$8+3n=8+3\cdot 3=17\\ 2+5m=2+5\cdot 3=17$$ which is the first common term. Hence: $$8+3n=8+3(3+5k)=17+15k$$ is the resulting common arithmetic series.