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What is the sum of the arithmetic sequence 136, 124, 112 …, if there are 36 terms?
–2,736
–2,664
–2,628
–2,700
 2 years ago
 2 years ago
What is the sum of the arithmetic sequence 136, 124, 112 …, if there are 36 terms? –2,736 –2,664 –2,628 –2,700
 2 years ago
 2 years ago

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mitosukiBest ResponseYou've already chosen the best response.0
I believe it is 2664?
 2 years ago

Mr.MathBest ResponseYou've already chosen the best response.2
What's the common difference between each successive members?
 2 years ago

Mr.MathBest ResponseYou've already chosen the best response.2
Oh sorry, I read the choices as the members of the sequence. My bad!
 2 years ago

ash2326Best ResponseYou've already chosen the best response.0
a1=136 an=a1+(n1)d an=136+(361)(12) an=136420 an=284 an=284 is the 36th term sum =n/2(a1+a36) =36/2(136284) =18*(148) 2664
 2 years ago
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