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find the sum of the integers between 2 and 100 which are divisible by 3

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    if 5 arithmetic means are inserted between 7 and 25 what is the middle mean to be inserted

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    The multiples of 3 between 2 and 100 are
    3, 6, 9, ... , 99 , an AS where a=3 and d=3

    how many of those are there?
    t(n) = a+(n-1)d
    99 = 3 + (n-1)(3)
    96 = 3n-3
    99 = 3n
    n = 33
    so now you want the sum of those 33 arithmetic terms
    S(33) = (33/2)(first + last) = (33/2)(3+99) = 1683

    Your second question....
    so your 7 becomes the first term, and your 25 becomes the 7th term
    25 = 7 + 6d
    d = 3
    so your middle term would be term(4)
    = a+3d = 7+9 = 16

    check: 7 10 13 16 19 22 25

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    1584

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