in a geometric series t1=23,t3=92 and the sum of all of the terms of the series is 62 813. how many terms are in the series?

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asked by cathy
  1. for this GS
    a = 23
    t(3) = ar^2
    92 = 23r^2
    r^2 = 4
    r = ± 2

    sum(n) = 62813

    sum(n) = a(r^n - 1)/(r-1)
    23(2^n -1)/1 = 62813
    2^n - 1 = 2731
    2^n = 2732

    n log2 = log2732
    n = 11.4

    something wrong here, n has to be a whole number.

    for 11 terms,
    sum(11) = 23(2^11 - 1)/1 = 47081
    for 12 terms
    sum(12) = 23(2^12 - 1)/1 = 94185
    there is no sum of 62813

    check your question or your typing
    Once you find your error, just change the corresponding numbers in my solution.

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    posted by Reiny
  2. Nice

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