Let the sum of the first $n$ terms of a non-constant A.P. $a_1,a_2,a_3,\dots \dots ..$be $50n+\frac{n(n-7)}{2}A$ where A is a constant. If $d$is the common difference of this A.P., then the ordered pair  $\left(d,a_{50}\right)$ is equal to

Given, $S_n=50n+\frac{n(n-7)A}{2}$

We know that $a_n=S_n-S_{n-1}$

$\begin{array}{l}=50n+1\frac{n(n-7)A}{2}-50(n-1)-\frac{(n-1)(n-8)A}{2}\\ =50+\frac{A}{2}\left[n^2-7n-n^2+9n-8\right]=50+A(n-4)\end{array}$

Now, $d=a_n-a_{n-1}=50+A(n-4)-50-A(n-5)=A$

And $a_{50}=50+(50-4)A=50+46A$

Hence, ordered pair $\left(d,a_{50}\right)=(A,50+46A)$