Can someone explain the reverse sum in arithmetic series?

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The discussion clarifies the concept of reversing sums in arithmetic series, particularly when summing numbers from 1 to 100. By pairing terms such as 1 + 100 and 2 + 99, it becomes evident that each pair sums to the same total, allowing for a simplified calculation. The formula sn = n/2 (2a + (n-1)d) is explained as a way to derive the total sum, where 'a' is the first term and 'd' is the common difference. The reversal method helps to visualize the consistent sum of pairs, reinforcing the relationship between the average of the series and the total number of terms. Understanding this concept is essential for grasping the derivation of the arithmetic series formula.
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Please can someone help me with arithmetic series.

I don't understand why you reverse the sum when summing 1 to 100.

also I don't understand the formula given : sn = n/2 ( 2a+(n-1)d )

thanks


Roger
 
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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.
 
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.
 
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)

That made it:

(1 + 100)/2 * 100 = 50.5 * 100 = 5050
average * number of data = sum
 
ooh that reverse order :) yes i remember that. its part of the derivation of the formula i think.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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