Question

Assuming all the terms of A.P $\left[a_n\right]$ are integers with $a _1=2019$ and for $n \in I ^{+}$these always exists positive integer $m$ such that $a_1+a_2+a_3+\ldots+a_{n-2}+a_{n-1}+a_n=a_m$, then find number of such sequences

Answer 1

Assuming common difference of $\left[a_n\right]$ is $d$ where $d \in I$
$\Rightarrow a_1+a_2=a_k$ for some positive integer $k$
$\Rightarrow 2 a_1+d=a_1+(k-1) d \Rightarrow a_1=(k-2) d$
$\Rightarrow k \neq 2$ and $d=\frac{a_1}{(k-2)}$ as $a_n=a_1+(n-1) d \Rightarrow a_n=a_1+\left(\frac{n-1}{n-2}\right) a_1$
$\Rightarrow a_1+a_2+a_3+\ldots+a_{n-2}+a_{n-1}+a_n=a_1 n+\frac{n(n-1)}{2} d$
$\Rightarrow \displaystyle\sum_{i=1}^n a_1=a_1+\left((n-1)(k-2)+\frac{n(n-1)}{2}\right) d$
As 2019 is product of points (3) and (673)
Hence k-2 $=-1,1,3,673,2019$