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Math Help - Find, in terms of k... [Arithmetic series]

  1. #1
    Junior Member Lonehwolf's Avatar
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    Find, in terms of k... [Arithmetic series]

    The first term of an arithmetic series is (5+3k) where k is a positive integer. The last term is 17(1+k) and the common difference is 2. Find, in terms of k,

    (i) the number of terms,
    (ii) the sum of the series.

    I'm stuck as early as the starting point, I'm trying to go over all of my notes but I can't seem to find anything that discusses first and last terms. I do recall having done these in the first year in college, but the notes are gone

    Can someone help me through this one please?
    Last edited by mr fantastic; May 5th 2009 at 05:30 AM. Reason: Removed quote tags around question
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by Lonehwolf View Post
    The first term of an arithmetic series is (5+3k) where k is a positive integer. The last term is 17(1+k) and the common difference is 2. Find, in terms of k,

    (i) the number of terms,
    (ii) the sum of the series.

    I'm stuck as early as the starting point, I'm trying to go over all of my notes but I can't seem to find anything that discusses first and last terms. I do recall having done these in the first year in college, but the notes are gone

    Can someone help me through this one please?
    t_n = t_1 + (n-1)d \Rightarrow 17(1 + k) = (5 + 3k) + 2(n-1).

    Solve for n in terms of k.

    Now use the usual formula for the sum of an arithmetic series.
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  3. #3
    MHF Contributor red_dog's Avatar
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    a_1=5+3k, \ a_n=17(1+k), \ r=2

    a_n=a_1+(n-1)r\Rightarrow 17+17k=5+3k+2n-2\Rightarrow n=7k+7

    S=\frac{(a_1+a_n)n}{2}

    Replace a_1, \ a_n, \ n and find S.
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  4. #4
    Junior Member Lonehwolf's Avatar
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    Thanks, the answers matched.
    Last edited by mr fantastic; May 6th 2009 at 04:33 AM.
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