# SOLUTION: prove that: Sum of n numbers in a sequence is n/2[2a+(n-1)d].

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 Algebra: Sequences of numbers, series and how to sum them Solvers Lessons Answers archive Quiz In Depth

 Question 104572: prove that: Sum of n numbers in a sequence is n/2[2a+(n-1)d].Found 2 solutions by stanbon, TP:Answer by stanbon(57214)   (Show Source): You can put this solution on YOUR website!prove that: Sum of n numbers in a sequence is n/2[2a+(n-1)d]. ---------------- Write the sequence : S(n)=a ,a+d ,a+2d ,.....a+(n-1)d Write the sequence backwards: S(n)=a+(n-1)d, a+(n-2d) ,a+(n-3d) ,a ----------------- Add the two sequences to get: 2S(n)= n[a+a+(n-1)d] 2S(n) = n[2a+(n-1)d] S(n) = (n/2)[2a+(n-1)d] ============ Cheers, Stan H. Answer by TP(29)   (Show Source): You can put this solution on YOUR website!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. If we write this in reverse we get S=l+(l-d)+(l-2d)+(l-3d)+....+(a+d)+a. So we have, S=a+(a+d)+(a+2d)+(a+3d)+....+(l-d)+l (i) and S=l+(l-d)+(l-2d)+(l-3d)+....+(a+d)+a (ii) Now add (i) and (ii) together, making sure that you add corresponding terms together and you get 2S=(a+l)+(a+l)+(a+l)....+(a+l)+(a+l). And so 2S= n(a+l) (because there are n lots of (a+l)) So S= [n(a+l)]/2 (iii) But your last term or nth term can be written as a+(n-1)d so l=a+(n-1)d Now replace l in (iii) and we get S=[n(a+a+(n-1)d)]/2 This simplifies to S=[n(2a+(n-1)d)]/2 Q.E.D. (N.B. l is the letter L and 1 is the number ONE)