SOLUTION: Find the sum of each of these series using the series formula. a) -11 - 3 + 5 + 13 + ... + 125 b) 6 - 18 + 54 - ... + 39366

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Question 1185789: Find the sum of each of these series using the series formula.
a) -11 - 3 + 5 + 13 + ... + 125
b) 6 - 18 + 54 - ... + 39366
Answer by ikleyn(52921) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find the sum of each of these series using the series formula.
a) -11 - 3 + 5 + 13 + ... + 125
b) 6 - 18 + 54 - ... + 39366
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Arithmetic progression with the first term of -11;  the last term of 125, and the common difference of 8.  
The number of terms n is


    n = %28125+-+%28-11%29%29%2F8 + 1 = %28125%2B11%29%2F8 + 1 = 136%2F8 + 1 = 17 + 1 = 18.


(the difference of the last term and the first term, divided by the common difference, PLUS 1).



Apply the formula for the sum of the first n terms of an arithmetic progression


    S%5Bn%5D = %28%28a%5B1%5D%2Ba%5Bn%5D%29%2F2%29%2An = %28%28-11%2B125%29%2F2%29%2A18 = %28114%2F2%29%2A18 = 57*18 = 1026.     ANSWER

Part  (a)  is solved.

Solve part  (b)  PRECISELY  in the same way, using my instructions.

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For introductory lessons on arithmetic progressions see
    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.


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