Answer to Question #21714 in Abstract Algebra for Samuel
Find the sum of 120 + 119 + 118 + .... + 3 + 2 + 1
1
2013-01-09T09:09:34-0500
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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