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The sum of n consecutive integers is 11. If the least of these integers is -10 what is the value of n ?
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Hint: The sum of the first n terms of any arithmetic sequence is Sn = (n*(2*a1 + d*(n-1)))/2 In this case, Sn = 11, a1 = -10 and d = 1, so plug this into the formula above to get 11 = (n*(2*(-10) + 1*(n-1)))/2 11*2 = n*(-20 + n-1) -20n + n^2-n = 22 Keep going to solve for n.
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