Let the sum of the first $$n$$ terms of a non-constant A.P., $${a_1},{a_2},{a_3},......\,\,{\text{be}}\,\,50n + \frac{{n\left( {n - 7} \right)}}{2}A,$$        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:

Solution :
$$\eqalign{ & \because \,\,{S_n} = \left( {50 - \frac{{7A}}{2}} \right)n + {n^2} \times \frac{A}{2} \cr & \Rightarrow \,\,{a_1} = 50 - 3A \cr & \therefore \,\,d = {a_2} - {a_1} = \left( {{S_2} - {S_1}} \right) - {S_1} \cr & \Rightarrow \,\,d = \frac{A}{2} \times 2 = A \cr & {\text{Now, }}{a_{50}} = {a_1} + 49 \times d \cr & = \left( {50 - 3A} \right) + 49A = 50 + 46A \cr & {\text{So,}}\,\left( {d,{a_{50}}} \right) = \left( {A,50 + 46A} \right) \cr} $$