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in a geometric series t1=23,t3=92 and the sum of all of the terms of the series is 6813. How many terms are in the series

  • math -

    t1 = 23 ---> a = 23
    t3 = 92 --->ar^2 = 92
    divide them
    r^2 = 4
    r = ± 2

    sum(n) = a(r^n - 1)/(r-1) = 6813

    If r = 2
    23( 2^n - 1)/(2-1) = 6813
    2^n - 1 = 296.21... ------> not a whole number
    so there can't be a sum of 6813
    if r = -2
    23((-2)^n - 1)/(-2-1) = 6813
    (-2)^n = -887.65...
    no way!

    This question either has a typo , or the question itself is flawed.


    proof:
    suppose we had 4 terms
    sum(4) = 23(2^4 - 1)/1 = 345
    sum(5) = 23(2^5 - 1)/1 = 713
    ...
    sum(8) = 5865 ----
    sum(9) = 11753

    or

    sequence is
    23 + 46 + 92 + 184 + 368 + 736 + 1472 + 2944 + 5888 +

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