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Answer to Question #21714 in Abstract Algebra for Samuel

Question #21714
Find the sum of 120 + 119 + 118 + .... + 3 + 2 + 1
Expert's answer
this is the sum of first 120 terms of arithmeticprogression so a1=120, a120=1, n=120 hence from the formula:
s=(120+1)*120/2=60*121=7260

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