SOLUTION: How many terms has an arithmetic {{{highlight(cross(progreression))}}} <U>progression</U> where sum is 1340 with first term 3 and nth term 131?

Question 1149896: How many terms has an arithmetic progression where sum is 1340 with first term 3 and nth term 131?
Answer by ikleyn(52879) (Show Source): You can put this solution on YOUR website!
.


For any arithmetic progression of n terms  , ,  . . . ,  the formula for its sum S works


    S = .


In our case, it gives


    1340 =  = .


Hence,


    n =  = 2*10 = 20.   


ANSWER.  The number of the terms in the progression is 20.

Solved.

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    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
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