Assuming all the terms of A.P. $[a_{n}]$ are integers with $a_{1}=2019$. Find number of such sequence satisfying given condition

Question

Assuming all the terms of A.P. $[a_{n}]$ are integers with $a_{1}=2019$ and for $n\in I^+$ there always exsits positive integer $m$ such that $a_{1}+a_{2}+.....+a_{n}=a_{m}$, then find number of such sequences

My Approach: Let say common difference of given A.P. is $d$.

Using sum of A.P. formula $a_{1}+a_{2}+...+a_{n}=\dfrac{n}{2}(2\cdot2019+(n-1)d)$

And Using formula of $n^{th}$ term of A.P. $a_m=2019+(m-1)d$

Now, according to question $\dfrac{n}{2}(4038+(n-1)d)=2019+(m-1)d$

$\implies n(4038+(n-1)d)=4038+2(m-1)d$

$\implies (n-1)4038+n(n-1)d=2(m-1)d$

$\implies \dfrac{(n-1)2019}{d}+\dfrac{n(n-1)}{2}=(m-1)$

Hence for psotive $m\in Integer, d$ must be a factor of $2019$ and factor of $2019$'s are $1,3,673,2019$ so total four such sequence exsit.

But given answer is $5$

Answer 1

While $m$ must be positive, $d$ might be negative. Consider the sequence with $d=-2019$; its values are 2019, 0, -2019, -4038, -6057, etc. The sums are 2019, 2019, 0, -2019, etc; these values are all in the sequence. Remaining sums are negative multiples of 2019, and all such are in the sequence.

Other negative values of $d$ would give negative values for $m$.