SOLUTION: How many terms must be added in an arithmetic sequence whose first term is 1919 and whose common difference is 66 to obtain a sum of 4012​?

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Question 1102883: How many terms must be added in an arithmetic sequence whose first term is 1919 and whose common difference is 66 to obtain a sum of 4012​?
Answer by josgarithmetic(39618) About Me  (Show Source):
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First term, 1919
Last term, 1919+66(n-1) for index n

The given sum 4012,
%28n%2F2%29%281919%2B1919%2B66%28n-1%29%29=4012
Simplify and solve.