SOLUTION: How many terms of the progression 3, 6, 9, 12,.... must be taken at the least to have a sum not less than 2000?

Algebra ->  Sequences-and-series -> SOLUTION: How many terms of the progression 3, 6, 9, 12,.... must be taken at the least to have a sum not less than 2000?      Log On


   

Question 727805: How many terms of the progression 3, 6, 9, 12,....
must be taken at the least to have a sum not less than 2000?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Finding or evaluating finite arithmetic series is %28n%2F2%29%282a%5B1%5D%2B%28n-1%29d%29, where n is the index of the final term, a%5B1%5D is the first term, and d is the common difference between terms.

Your example would use the expression for the sum as %28n%2F2%29%282%2A3%2B%28n-1%29%2A3%29, and because you want this value to be "not less than 2000", you want the inequality statement,
highlight%28%28n%2F2%29%282%2A3%2B%28n-1%29%2A3%29%3E=2000%29. Solve for n.