Let a, b and c be the 7th’ 11th and 13th terms respectively of a non-constant A.P, If these are alsothe three consecutive terms of a G.P., then ac=

Let a, b and c be the 7th' 11th and 13th terms respectively of a non-constant A.P, If these are also

the three consecutive terms of a G.P., then ac=

  1. A

    2

  2. B

    1/2

     

  3. C

    4

  4. D

    7/13

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    Solution:

    We have, a=A+6d;b=A+10d;c=A+12d

    Since, a, b and  c are in G.P .

        (A+10d)2=(A+6d)(A+12d)    A2+100d22+20Ad=A2+18Ad+72d2

     2Ad=28d2Ad=14

    [d0 as A.P. is non constant]

    Now, ac=A+6dA+12d=Ad+6Ad+12=14+614+12=4

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