How many terms do the two sequences $S_1$ and $S_2$ have in common?
$S_1 = 1, 3, 6, 10, 15\dots$ up to $200$ terms.
$S_2 = 3, 6, 9, 12, 15\dots$ up to $200$ terms.
I need to know the number of common terms in these two sequences (irrespective of their position in respective series.)
For example, $3,6$ and $15$ are a few of the common terms in these two sequences, but I need to know the exact number of common terms.
My approach:
$n^{th}$ term of $S_1 = \dfrac{[n(n+1)]}{2}$
$k^{th}$ term of $S_2 = 3k$
Now for common terms:
$\dfrac{[n(n+1)]}{2} = 3k$
Which gives:
$n(n+1) = 6k$
Had the above equation been linear in terms of $n$ and $k$ then I might have been able to solve it for possible integral values of $k$ and $n$ but now I am stuck here and don't know how to proceed further. Kindly provide assistance in solving this further.
Answer is 22 common terms
Best Answer
The question is: For how many $n\ge1$ is $\frac{n(n+1)}{2}$ a multiple of $3$ and $\le 600$?
We first check that $\frac{n(n+1)}{2}\le 600$ iff $n\le 34$ (this is because $\lfloor\sqrt{1200}\rfloor = 34$). Next, the integer $\frac{n(n+1)}{2}$ is a multiple of $3$ iff $n$ or $n+1$ is a multiple of $3$, that is unless $n\equiv 1\pmod 3$. Thus two thirds of the $33$ numbers $n=1,\ldots, 33$ lead to a common term. The remaining $n=34$ is $\equiv 1\pmod 3$, so does not add another solution. Thus the answer is $22$.