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?

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) (Show Source): You can put this solution on YOUR website!
First term, 1919
Last term, 1919+66(n-1) for index n

The given sum 4012,

Simplify and solve.
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