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Thread: Arithmetic Sequence

  1. #1
    Junior Member
    Sep 2007
    Oregon, West Coast ^_^

    Arithmetic Sequence

    1. Find the first two terms of an arithmetic sequence where the sixth term is 21 and the sum of the first seventeen terms is 0.

    2. Five consecutive terms of an arithmetic sequence have a sum of 40. The product of the middle and two ends terms is 224. Find the terms of the sequence.
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  2. #2
    MHF Contributor
    Aug 2007
    #1 - You may not know this, but one of the reasons for the study of mathematics is to train you to think in a way with which you are not familiar. Everyone thinks a little differently, and that's fine. If everyone thought like I do, the world would be a very strange place. It is not to make you forget how you usually think. It is to show you that there is another important way to proceed. Mathematical thinking requires perhaps more order than some normally would use. Order is important.

    Having said that, let's think through these problems.

    We have an arithmetic sequence. What do we know about those?

    It has a first term, usually called 'a'.
    It has a common difference, usually called 'r'.
    The first term, then, is 'a'.
    The second term, then, is 'a+r'.
    The third term is a + 2r
    The fourth term is a + 3r
    The 10th term is a + 9r
    The 100th term is a + 99r.

    Are you seeing a pattern?

    I think the nth term is a + (n-1)r -- Do you agree?

    Okay, that is what we know about arithmetic sequences. Now, what was the question?

    Find the first two terms,

    Okay, we know already that is a and a + r. Find a and r and we're done.

    What information have we?

    The 6th term is 21

    Fine, then a + 5r = 21

    the sum of the first seventeen terms is 0

    Okay, adding up terms is a little trickier than just naming terms, but it's nto too bad.

    Let's go back into Exploration Mode

    Sum of the first term is 'a'.
    Sum of the first two terms is (a) + (a+r) = 2a + r.
    Sum of the first three terms is (a) + (a+r) + (a+2r) = 3a + 3r.
    Sum of the first four terms is (a) + (a+r) + (a+2r) + (a+3r) = 4a + 6r.

    Are you seeing the pattern? The 'a'-part is easy. It's the 'r'=part that is a little sticky. However, and I'll spare you the development, these are well-known numbers. 1, (1+2), (1+2+3), (1+2+3+4), (1+2+3+4+5). These are the triangular numbers and have a known simple formula. If there are n terms, the sum is (n)(n+1)

    Thus, the sum of the first n terms is n*a + (n-1)(n)*r

    You tell me why that did not appear like you expected. It is your homework for the day.

    Okay, what were we doing?

    the sum of the first seventeen terms is 0

    Right, then 17*a + (16)(17)*r = 0 = 17a + 136r

    Now we have only an algebra problem. Two equations in two unknowns.

    a + 5r = 21
    17a + 136r = 0

    Solve that for a and r and you are nearly done.

    Note: Is is rather difficult to write an entire thought process on a very small space. The point is that you need to be orderly. You need to switch gears when you need to. There are just a few primary modes where your mind must travel:

    What does it want?
    What does it offer?
    What do I know?
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