JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • 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) \[\left( 50,50+45A \right)\]

    Correct Answer: A

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner