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The third term of a sequence is 2(2/3) and the sixth term is 9. Find the first six terms when the sequence is; (a) arithmetic (b) geometric

  • Maths -

    a) for arithmetic:
    t3 = a+2d = 4/3
    t6 = a+5d = 9
    subtract them
    3d = 23/3
    d = 23/9
    then a+2(23/9) = 4/3
    a = -34/9

    first 6 terms are:
    -34/9, -11/9 , 4/3 , 35/9 , 58/9 , 9

    well, that checked out ok


    b) for geometric
    t3 = ar^2 = 4/3
    t6 = ar^5 = 9
    divide them
    r^3 = 27/4 = 54/8 , so I can take a nice cube root of the bottom
    r = (3/2)(2^(1/3)
    then a(9/4)(2^(2/3) = 4/3
    a = (16/27) 2^(-2/3)

    messy subs , you try them

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