Arithmetic

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Find a geometric series which has a third term of 7/4 and a sum to infinity of 14.

  • Arithmetic -

    ar^2 = 7/4

    a/(1-r) = 14
    a = 14(1-r)

    back into the 1st...

    14(1-r)r^2 = 7/4
    8(1-r)r^2 = 1
    8r^2 - 8r^3 - 1 = 0
    8r^3 - 8r^2 + 1 = 0

    after a few trial-and-error attempts, I found
    f(1/2) = 0 ,so 2r-1 was a factor

    (2r-1)(4r^2 -2r-1) = 0

    r = 1/2 or r = (1 ± √5)/4

    if r = 1/2 , then a = 14(1/2) = 7
    GS is 7, 7/2, 7/4 , 7,8 , ....

    if r = (1 + √5)/4
    we have a contradiction with the sum of the infinitite series
    we can only have such a sum if |r| < 1
    and both of the irrational answers fall outside of that domain.

    so the GS is 7, 7/2, 7/4, ....

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