Let the sum of the first n terms of a non-constant A.P. a1,a2,a3,…….. be 50n+n(n−7)2A where A is a constant. If d is the common difference of this A.P., then the ordered pair  d,a50 is equal to

Let the sum of the first n terms of a non-constant A.P. a1,a2,a3,.. be 50n+n(n7)2A where A is a constant. If d is the common difference of this A.P., then the ordered pair  d,a50 is equal to

  1. A

    (A, 50 + 45A)

  2. B

    (A, 55+ 45A)

  3. C

    (50, 50 + 46A)

  4. D

    (50, 50 + 45A)

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    Solution:

    Given, Sn=50n+n(n7)A2

    We know that an=SnSn1

    =50n+1n(n7)A250(n1)(n1)(n8)A2=50+A2n27nn2+9n8=50+A(n4)

    Now, d=anan1=50+A(n4)50A(n5)=A

    And a50=50+(504)A=50+46A

    Hence, ordered pair d,a50=(A,50+46A)

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