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



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mitosuki
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I believe it is 2664?

Mr.Math
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Are you sure?

Mr.Math
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What's the common difference between each successive members?

Mr.Math
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Oh sorry, I read the choices as the members of the sequence. My bad!

ash2326
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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

Mr.Math
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2664 is right.

mitosuki
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Haha thank you :)

Mr.Math
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Haha you're welcome :P