document.write( "Question 104572: prove that:
\n" ); document.write( " Sum of n numbers in a sequence is n/2[2a+(n-1)d]. \n" ); document.write( "
Algebra.Com's Answer #76102 by TP(29)\"\" \"About 
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The general series is S=a+(a+d)+(a+2d)+(a+3d) +....+(l-d))+l where a is the first term,d is the common difference and l is the last term.\r
\n" ); document.write( "\n" ); document.write( "If we write this in reverse we get S=l+(l-d)+(l-2d)+(l-3d)+....+(a+d)+a.\r
\n" ); document.write( "\n" ); document.write( "So we have,
\n" ); document.write( "S=a+(a+d)+(a+2d)+(a+3d)+....+(l-d)+l (i)
\n" ); document.write( "and
\n" ); document.write( "S=l+(l-d)+(l-2d)+(l-3d)+....+(a+d)+a (ii)\r
\n" ); document.write( "\n" ); document.write( "Now add (i) and (ii) together, making sure that you add corresponding terms together and you get
\n" ); document.write( "2S=(a+l)+(a+l)+(a+l)....+(a+l)+(a+l).
\n" ); document.write( "And so
\n" ); document.write( " 2S= n(a+l) (because there are n lots of (a+l))
\n" ); document.write( "So
\n" ); document.write( "S= [n(a+l)]/2 (iii)\r
\n" ); document.write( "\n" ); document.write( "But your last term or nth term can be written as a+(n-1)d so
\n" ); document.write( "l=a+(n-1)d
\n" ); document.write( "Now replace l in (iii) and we get
\n" ); document.write( " S=[n(a+a+(n-1)d)]/2
\n" ); document.write( "This simplifies to \r
\n" ); document.write( "\n" ); document.write( " S=[n(2a+(n-1)d)]/2 Q.E.D.\r
\n" ); document.write( "\n" ); document.write( "(N.B. l is the letter L and 1 is the number ONE)\r
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