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whats the sum of the sequence 1 2 3 4 5... 99?

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fundementals of math

Oct 18th, 2017

1 2 3 4 5 6 7 8 9..................99

first term = a=1

difference= d=2-1=3-2=4-3=1

number of terms= n = 99

S99= n/2[2a+(n-1)d]

       = 99/2[2*1+(99-1)1]

       = 99/2[2+98*1]

       = 99/2[2+98]

       = 99/2*100

       =99*50

       = 4950

Oct 13th, 2014

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