# Arithmetic series

1. Oct 28, 2004

### roger

Please can someone help me with arithmetic series.

I dont understand why you reverse the sum when summing 1 to 100.

also I dont understand the formula given : sn = n/2 ( 2a+(n-1)d )

thanks

Roger

2. Oct 28, 2004

### T@P

what do you mean by reverse the sum...? when adding the numebrs 1-100, its is clear that 1 + 100 = 101, 2 + 99 = 101, etc
therefore you can notice that you would take 101 and multiply by the number of terms. however, this would give you 2 times the sum because 2 + 99 = 99 + 2... so you then divide by 2. This in part explains the formula you have, and also notice that a + (n-1) * d is really an expression for the last term in terms of the first and d... if your looking for a complete derivation of the formula(s) look in any algebra 2 textbook.

3. Oct 28, 2004

### HallsofIvy

Exacly as T@p said. If you wanted to add from, say 1+ 2+ 3+ 4+ 5+ 6 (yes, I know that's easy to do directly- its a simple example) you could write
1+ 2+ 3+ 4+ 5+ 6 and reverse:
6+ 5+ 4+ 3+ 2+ 1 and notice that the sum of each pair of numbers is 7!
________________________
7+ 7+ 7+ 7+ 7+ 7

since there are 6 numbers in the original sum there are 6 7's: a total of 42. But since we have added 1+ 2...+ 6 TWICE (once in reversed order), we have to divide this by 2: 42/2= 21 which is, in fact, is 1+2+3+4+5+6.

The PURPOSE of reversing the order was to get the pairs of numbers all giving the same sum. In a general arithmetic series, going from a<sub>n</sub> to a<sub>n+1</sup> we ADD the "common difference" d. When we reverse the order, we are now SUBTRACTING d: the "+d" and "-d" cancel so we always get the same sum of pairs.

4. Oct 28, 2004

### Alkatran

Personnaly when I was presented this problem for the first time I:

-Realized that the sum of all these terms would be the same as the average of all these terms times the number of terms
-Since it's a line, the average is right at the center (or [1 + 100]/2)