Let $a_1, a_2, a_3,\dots ,a_n\dots$ be an A.P. whose common difference is an integer and $S_n$ denote the sum of its $n$ terms . If $a_1=1,a_n=300$ and $15\le n\le 50,$ then the ordered pair $\left(S_{n-4}, a_{n-4}\right)$ is equal to 

Let d be the common difference of the given A.P. Then $a_n=300$

$\begin{array}{l}\Rightarrow a_1+(n-1)d=300\\ \Rightarrow 1+(n-1)d=300\Rightarrow d=\frac{299}{n-1}=\frac{13\times 23}{n-1}\end{array}$

It is given that d is an integer. 

$\begin{array}{r}\therefore n-1=\pm 1, \pm 13, \pm 23, \pm 299\\ \Rightarrow n=0, 2,-12, 14,-22, 24,-298, 300\end{array}$

$\Rightarrow n=24 [\because 15\le n\le 50]$

$\therefore d=\frac{13\times 23}{n-1}\Rightarrow d=13.$

Thus,

          $a_{n-4}=a_{20}=a_1+19d=1+19\times 13=248$

and   $S_{1-4}=S_{20}=\frac{20}{2}\left(a_1+a_{20}\right)=10(1+248)=2490$

Hence, $\left(S_{n-4}, a_{n-4}\right)=(2490, 248)$