Let the sum of the first n terms of a non-constant A.P.,a_{1},a_{2},a_{3},..... be 50n+\frac{n\left ( n-7 \right )}{2}A, whrere 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 :

  • Option 1)

    \left ( 50,50+46A \right )

  • Option 2)

    \left ( 50,50+45A \right )

  • Option 3)

     \left ( A,50+45A \right )

  • Option 4)

    \left ( A,50+46A \right )

 

Answers (1)

Given ,S_{n}=50n+\frac{n\left ( n-1 \right )}{2}A...................(1)

  S_{n-1}=50_{\left ( n-1 \right )}+\frac{\left ( n-1 \right )\left ( n-2 \right )}{2}A.........................(2)

T_{n}=S_{n}-S_{n-1}=50+A\left ( n-4 \right )

T_{1}=50+\left ( -3A \right )

T_{2}=50+\left ( -2A \right )

d=A

T_{50}=50-3A+\left ( 50-1 \right )A 

        =50+46A


Option 1)

\left ( 50,50+46A \right )

Option 2)

\left ( 50,50+45A \right )

Option 3)

 \left ( A,50+45A \right )

Option 4)

\left ( A,50+46A \right )

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