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    The sum to infinity of a geometric series is 5 and the first term is 2.
    Find the common ratio of the series..

    and The first 3 terms of an arithmetical progression are 7, 5.9 and 4.8

    Find
    i)the common difference
    ii)the smallest value of n for which the sum to n terms is negative
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    2/(1 - r) = 5
    1 - r = 2/5
    r = 3/5

    --

    d = 5.9 - 7 = -1.1

    S(n)
    = sum of first n terms
    = (1/2)n[first term + nth term]
    = (1/2)n[7 + (7 - 1.1(n - 1))]
    = (1/2)n[14 - 1.1(n - 1)]

    S(n) is negative when

    14 - 1.1(n - 1) < 0
    (n - 1) > 14/1.1
    n > 14/1.1 + 1

    Smallest such value of n = 14
 
 
 
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Updated: January 5, 2005

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