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mitosuki

  • 4 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

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  1. mitosuki
    • 4 years ago
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    I believe it is -2664?

  2. Mr.Math
    • 4 years ago
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    Are you sure?

  3. Mr.Math
    • 4 years ago
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    What's the common difference between each successive members?

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

  5. ash2326
    • 4 years ago
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    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

  6. Mr.Math
    • 4 years ago
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    -2664 is right.

  7. mitosuki
    • 4 years ago
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    Haha thank you :)

  8. Mr.Math
    • 4 years ago
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    Haha you're welcome :P

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