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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

Mathematics
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I believe it is -2664?
Are you sure?
What's the common difference between each successive members?

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Other answers:

Oh sorry, I read the choices as the members of the sequence. My bad!
a1=136 an=a1+(n-1)d an=136+(36-1)(-12) an=136-420 an=-284 an=-284 is the 36th term sum =n/2(a1+a36) =36/2(136-284) =18*(-148) -2664
-2664 is right.
Haha thank you :)
Haha you're welcome :P

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