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the third term of a geometric sequence is 6 and the sixth term is 3 over 32. Determine the first terms and the general term Tn?

  • maths -

    1 a
    2 ar
    3 ar^2
    4 ar^3
    5 ar^4
    6 ar^5

    n ar^(n-1)

    so
    ar^2 = 6
    ar^5 = 3/32
    so
    a = 6/r^2
    (6/r^2)r^5 = 3/32
    6/r^3 = 3/32
    r^3 = 32*2 = 64
    r = 4
    then
    a = 6/16 = 3/4
    so
    Tn = (3/4)4^(n-1)

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