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

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(75887) (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].
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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
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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]
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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)

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