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Math Help - Two series/sequence problems

  1. #1
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    Two series/sequence problems

    Evening all, two questions here, first off I'm just sceptical with my answer for the first question, in great need of a second opinion in terms of my methods etc. The second question I'm really struggling where to start. Help would be much appreciated!

    Firstly: The sum of the first four terms of an arithmetic sequence is 139 and the sum of the next four terms is 115. Find a and d:

    [A] 4/2(2a + 3d) = 139
    4a + 6d = 139

    [B] 8/2(2a + 7d) = 115
    8a + 28d = 115

    Solving simultaneously: 16d = -163
    d = -163/16

    Substituting into equation [A]: 4a + 6(-163/16)=139
    a = 1601/32




    Second problem:

    Given that, Un+1 = 3 - 1/3Un and U1 = 3

    a) Find the values for U2, U3 and U4
    b) Find the limiting value of Un as n tends to infinity.
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  2. #2
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    Re: Two series/sequence problems

    For the second one, let L be the limiting value, hence the recursive definition will give you:

    L=3-\frac{L}{3}

    Solve for L.
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    Re: Two series/sequence problems

    Quote Originally Posted by DonGorgon View Post
    Firstly: The sum of the first four terms of an arithmetic sequence is 139 and the sum of the next four terms is 115. Find a and d:
    [A] 4/2(2a + 3d) = 139
    4a + 6d = 139

    [B] 8/2(2a + 7d) = 115
    8a + 28d = 115
    You have a mistake in the second equation.
    It should be
    \frac{8(2a + 7d)}{2}=139+115=254

    Because it says "sum of the next four terms is 115"
    It is the sum of first four plus the sum of the next four terms give the eight.
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    Re: Two series/sequence problems

    Quote Originally Posted by Plato View Post
    You have a mistake in the second equation.
    It should be
    \frac{8(2a + 7d)}{2}=139+115=254

    Because it says "sum of the next four terms is 115"
    It is the sum of first four plus the sum of the next four terms give the eight.

    Ah thanks, so from that I have:

    16d = -24
    d = -3/2

    Therefore, [A] 8a -18 = 278

    a = 296/8 = 37
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    Re: Two series/sequence problems

    In terms of the series with d = -3/2 and a =37. how would you find the value for n, for which the nth term is the last positive term in the series?
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    Re: Two series/sequence problems

    Quote Originally Posted by DonGorgon View Post
    In terms of the series with d = -3/2 and a =37. how would you find the value for n, for which the nth term is the last positive term in the series?
    Can you solve 37-1.5(n-1)<0~?
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