Trevor Pythagoras

An arithmetic series or progression is one in which each term is the previous plus a fixed amount eg)
2 + 5 + 8 + 11 + 14 + 17 + … where each term is 3 more than the previous
In general the nth term in an arithmetic series is
T_n = a + (n-1)d
where a is the first term in the series (2 in the above example) and d is the difference between the terms (3 in the example). We can use this to calculate any term in the series without having to work out all of the previous terms, for example we can calculate the 100th term in the example to be 2 + 99×3 = 299. This formula works because the nth term is equal to the (n-1)th term plus d which in term is equal to the (n-2)th term until n-1 d’s have been added to the initial term a.

In the example i have written the series as
2 + 5 + 8 + 11 + 14 + 17 + …
instead of
2 , 5 , 8 , 11 , 14 , 17, …
because it is often interesting to look at sum of n items in a series. The sum S_n of the first n terms in a series can be found using
S_n = n(2a+(n-1)d)/2

Proof of sum of series
We can show that
S_n = n(2a+(n-1)d)/2
as follows

Write out the series in order and then in reverse order as follows
S_n = a + (a+d) + (a+2d) + …. + (a-(n-1)d) + (a+nd)
S_n = (a+(n-1)d) + (a+(n-2)d) + (a+(n-3)d) + …. + (a+d) + a
Now if we add each of the terms in the two sums we get
2S_n = (2a+(n-1)d) + (2a+(n-1)d) + (2a+(n-1)d) + … + (2a+(n-1)d) + (2a+(n-1)d)

where the 2a+(n-1)d repeatsn times. This is because we adding the rth term with the(n-r)th term hence we doing
(a+(r-1)d)+(a+(n-r)d)=2a+(n-1)d
so
2S_n=n(2a+(n-1)d)
therefore
S_n=n(2a+(n-1)d)/2