1. Given sum of first $n$ terms

   $S_n = 50n + \frac{n(n-7)}{2}A$

   We need to find the common difference $d$ and the $50^{\text{th}}$ term $a_{50}$.

2. Find the general term using $a_n = S_n - S_{n-1}$

   First write:

   $S_{n-1} = 50(n-1) + \frac{(n-1)(n-8)}{2}A$

   Therefore,

   $a_n = S_n - S_{n-1}$

   $a_n = \left(50n + \frac{n(n-7)}{2}A\right) - \left(50(n-1) + \frac{(n-1)(n-8)}{2}A\right)$

   $a_n = 50 + \frac{1}{2}A\left[n(n-7) - (n-1)(n-8)\right]$

3. Simplify the bracket

   $n(n-7) = n^2 - 7n$

   $(n-1)(n-8) = n^2 - 9n + 8$

   So,

   $n(n-7) - (n-1)(n-8) = (n^2-7n) - (n^2-9n+8) = 2n-8$

   Hence,

   $a_n = 50 + \frac{1}{2}A(2n-8)$

   $a_n = 50 + (n-4)A$

4. Compare with the standard A.P. form

   For an A.P.,

   $a_n = a_1 + (n-1)d$

   We have

   $a_n = 50 + (n-4)A = (50-3A) + (n-1)A$

   Therefore,

   $d = A$

5. Find $a_{50}$

   $a_{50} = 50 + (50-4)A = 50 + 46A$

6. Ordered pair

   $(d, a_{50}) = (A, 50+46A)$

7. Match with options

   This corresponds to Option C.

Comparison with stored correct answer:

Stored correct answer is C, which matches our derived answer.