• # question_answer Let the sum of the first n terms of a non constant $A.P.,{{a}_{1}},{{a}_{2}},{{a}_{3}},$..... 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 $(d,{{a}_{50}})$ is equal to [JEE Main 9-4-2019 Morning] A) $\left( A,50+46A \right)$B) $\left( A,50+45A \right)$C) $\left( 50,50+46A \right)$      D) $\left( 50,50+45A \right)$

${{S}_{n}}=50n+\frac{n(n-7)}{2}A$             ${{T}_{n}}={{S}_{n}}-{{S}_{n-1}}$             $=50n+\frac{n(n-7)}{2}A-50(n-1)-\frac{(n-1)(n-8)}{2}A$             $=50+\frac{A}{2}[{{n}^{2}}-7n-{{n}^{2}}+9n-8]$             $=50+A(n-4)$             $d={{T}_{n}}-{{T}_{n-1}}$             $=50+A(n-4)-50-A(n-5)$             $=A$             ${{T}_{50}}=50+46A$             $(d,{{A}_{50}})=(A,50+46A)$