To find out the sum of arithmetic progression it has been given S=n(a+l)/2 And this formula has been derived in this manner $$ S=a+(a+d)+(a+2d)+.........+(l-2d)+(l-d)+l $$ And by reversing $$ S=l+(l-d)+(l-2d)+........+(a+2d)+(a+d)+a $$ On adding both we get $$ 2S=(a+l)+(a+l)+(a+l)+..........\text{ up to $n$ terms} $$ $$ 2S=n(a+l) $$ $$ S=n(a+l)/2 $$ My question is that why in deriving this formula reversing of the sum takes place(that is) $$ S=l+(l-d)+(l-2d)+........+(a+2d)+(a+d)+a $$ Why not any other method?
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Jun 17 '19 at 09:26
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You can also prove the formula by induction, but the method you mentioned is very elegant and easy to understand. Why should we despense with such a beautiful method ? – Peter Jun 17 '19 at 09:32
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1Why not this method, the simplest one method? – Grešnik Jun 17 '19 at 09:33
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Yes simple but reason behind doing this – user230507 Jun 18 '19 at 04:40