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sum of the first n integers

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1 Answer

an = a1+(n-1)d ... where an is the nth term, a1 is the first term, n is the number of terms, and d is the difference between terms in the sequence.
 
Sum (n) = n*(a1+an)/2
 
so ...
 
a) 303 = 3 + (n-1)(3) ... 303 = 3+3n-3 ... 3n = 303 ... n = 101 ... sum = 101*(3+303)/2 = 15453
 
b) 102 = 6+(n-1)(1) ... 102=6n-6 ... 6n = 108 ... n = 18 ... sum = 18*(6+102)/2 = 972
 
 

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