Can someone explain the reverse sum in arithmetic series?

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    Arithmetic Series
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Discussion Overview

The discussion revolves around understanding the concept of reversing the sum in arithmetic series, specifically in the context of summing the integers from 1 to 100. Participants explore the reasoning behind this method and the associated formula for the sum of an arithmetic series.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Roger expresses confusion about the concept of reversing the sum and the formula for the sum of an arithmetic series, sn = n/2 (2a + (n-1)d).
  • One participant explains that reversing the sum allows for pairing numbers (e.g., 1 + 100, 2 + 99) to simplify the calculation, noting that this results in pairs that sum to the same value.
  • Another participant illustrates the concept using a simple example of summing numbers from 1 to 6, demonstrating how reversing the order leads to consistent pair sums and the necessity of dividing by 2 to avoid double counting.
  • A different participant reflects on their initial understanding, stating that the sum can be calculated as the average of the terms multiplied by the number of terms, highlighting the average's position at the center of the series.
  • One participant acknowledges the reverse order as part of the derivation of the formula for the sum of an arithmetic series.

Areas of Agreement / Disagreement

Participants present various explanations and methods for understanding the reverse sum in arithmetic series, but there is no consensus on a single explanation or approach. The discussion remains exploratory with multiple perspectives offered.

Contextual Notes

Some participants reference the formula for the sum of an arithmetic series without fully deriving it, and there are assumptions about the understanding of basic arithmetic series concepts that may not be explicitly stated.

roger
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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.
 

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