If the  ratio of the  sum  of n terms of two  AP’s be (7n  + 1):(4n + 27),  then the  ratio of their 11th  terms will be

If the  ratio of the  sum  of n terms of two  AP's be (7n  + 1):(4n + 27),  then the  ratio of their 11th  terms will be

  1. A

    2:3

  2. B

    3:4

  3. C

    4:3

  4. D

    5:6

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    Solution:

    Let Sn, and S' be the sums of n terms of two AP's and T11 and T11, be the respective 11th term, then

    SnSn=n2[2a+(n1)d]n22a+(n1)d=7n+14n+27 (given)

    a+(n1)2da+(n1)2d=7n+14n+27 

    Now, put n=21, we get 

    a+10da+10d=T11T11=148111=43

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