Let $1,4,...$ and $9,14,...$ be two arithmetic progressions. Then the number of distinct integers in the collection of first $500$ terms of each of progressions is $$A. 833,\space B. 835,\space C.837 ,\space D. 901$$
My solution goes like this:
The two A.P.'s general term might be represented as $$1+(n_1-1)3=9+(n_2-1)5\implies 3n_1-5n_2=6\implies 3(n_1-2)=5n_2.$$ Hence, $3|n_2\implies n_2=3k$ and so, $3(n_1-2)=5n_2=15k\implies n_1-2=5k\implies n_1=5k+2.$ Now, $5k+2\leq 500\implies 5k\leq 498\implies k\leq 99.$ Due to this, we may conclude, $99$ terms of both the A.P.'s are similar. So, total number of common terms in both the A.P's is : $99+99=198$ and the number of distinct terms in both the A.P's in total is: $1000-198=802.$
However, I feel something's wrong with my solution, as my solution, is not matching with any of the options. But I don’t understand where is the solution going wrong, if that's the case. I am not quite getting it...