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Thread: Finding the summation of a converged arithmetic series: Two Answers

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    Senior Member x3bnm's Avatar
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    Finding the summation of a converged arithmetic series: Two Answers

    Is it possible to kindly check if I am right about finding the sum of a converged arithmetic series? Also the two answers?

    Thank you.

    deriving-summation-formula.pdf
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    Re: Finding the summation of a converged arithmetic series: Two Answers

    "Cannot display pdf file" is what was displayed to me when I clicked that.
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    Re: Finding the summation of a converged arithmetic series: Two Answers

    Quote Originally Posted by greg1313 View Post
    "Cannot display pdf file" is what was displayed to me when I clicked that.
    Here is what is contains:
    Sn = (a + (n − 1)d) + (a + (n − 2)d) + (a + (n − 3)d)... + (a + 2d) + (a + d) + a Sn = a + (a + d) + (a + 2d) + ... + (a + (n − 3)d) + (a + (n − 2)d) + (a + (n − 1)d) 2Sn = 2na + [(n − 1)d + d] + [(n − 2)d + 2d] + ... + [2d + (n − 2)d] + [d + (n − 1)d] 2Sn = 2na + (n − 1)nd Sn = n 2 [2a + (n − 1)d] or Sn = n 2 [a + (a + (n − 1)d)] = n 2 (a + l) or n 2 (a0 + an)..........[Answer]

    I'll be damn if I know what it means or what it is supposed to do??
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    Senior Member x3bnm's Avatar
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    Re: Finding the summation of a converged arithmetic series: Two Answers

    Quote Originally Posted by Plato View Post
    Here is what is contains:
    Sn = (a + (n − 1)d) + (a + (n − 2)d) + (a + (n − 3)d)... + (a + 2d) + (a + d) + a Sn = a + (a + d) + (a + 2d) + ... + (a + (n − 3)d) + (a + (n − 2)d) + (a + (n − 1)d) 2Sn = 2na + [(n − 1)d + d] + [(n − 2)d + 2d] + ... + [2d + (n − 2)d] + [d + (n − 1)d] 2Sn = 2na + (n − 1)nd Sn = n 2 [2a + (n − 1)d] or Sn = n 2 [a + (a + (n − 1)d)] = n 2 (a + l) or n 2 (a0 + an)..........[Answer]

    I'll be damn if I know what it means or what it is supposed to do??
    Sorry for misunderstanding.

    $\displaystyle \begin{align*}&S_{n} = (a + (n-1)d) + (a + (n-2)d) + (a + (n-3)d)... + (a + 2d )+ (a + d)\\& + a \\&S_{n} = a + (a + d) + (a + 2d ) + ... + (a + (n-3)d) + (a + (n-2)d)\\& + (a+ (n-1)d)\\&2S_{n} = 2na + [(n-1)d + d] + [(n-2)d + 2d] + ...+ [2d + (n-2)d]\\& + [d + (n-1)d]\\&2S_{n} = 2na + (n-1)nd\\&S_{n} = \frac{n}{2}[2a + (n-1)d]\\\text{or }S_{n} &= \frac{n}{2}[a + (a+ (n-1)d)]\\&=\frac{n}{2}(a+l)\text{ or } \frac{n}{2}(a_{0}+a_{n})\text{..........[Answer]}\end{align*}$

    first line is the summation from nth term or l to first term a, and Second line is the summation of n terms from first term a, to nth terms then I added them and and then simplified it.
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    Junior Member Walagaster's Avatar
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    Re: Finding the summation of a converged arithmetic series: Two Answers

    These are standard formulas for the sum of an arithmetic progression. I'm not sure what your question is but you can find many resources on the internet which discuss this. One that pops up is http://www.mathcentre.ac.uk/resource...pgp-2009-1.pdf
    Look at pages 4-6 of that.
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    Senior Member x3bnm's Avatar
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    Re: Finding the summation of a converged arithmetic series: Two Answers

    Quote Originally Posted by Walagaster View Post
    These are standard formulas for the sum of an arithmetic progression. I'm not sure what your question is but you can find many resources on the internet which discuss this. One that pops up is http://www.mathcentre.ac.uk/resource...pgp-2009-1.pdf
    Look at pages 4-6 of that.
    Thank you. Yes that was what I was looking for. Again thank you.
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