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
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
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.
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.