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superhelp101

  • one year ago

what is the sum of first 150 terms in sequence of..

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  1. superhelp101
    • one year ago
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    |dw:1432790538670:dw|

  2. superhelp101
    • one year ago
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    a) 61200 b)90000 c)122400 d)180000

  3. Nnesha
    • one year ago
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    it's arithmetic seq so formula is \[\huge\rm s_n = (\frac{ a_1 +a_n }{ 2 })\] a_n = last term

  4. superhelp101
    • one year ago
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    how do i find the last term?

  5. rational
    • one year ago
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    you may use \(n\)th term formula \[\large a_n = a_1 + (n-1)d\]

  6. Nnesha
    • one year ago
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    plug in 150 for n

  7. superhelp101
    • one year ago
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    a sub 149 -8

  8. rational
    • one year ago
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    |dw:1432791026176:dw|

  9. Nnesha
    • one year ago
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    correction \[\huge\rm s_n =\color{red}{n} (\frac{ a_1 +a_n }{ 2 })\] a_n = last term

  10. superhelp101
    • one year ago
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    so |dw:1432791117579:dw|

  11. rational
    • one year ago
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    Yes, find \(a_n\) and plug it in

  12. superhelp101
    • one year ago
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    still don't know how

  13. superhelp101
    • one year ago
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    wait..

  14. rational
    • one year ago
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    http://gyazo.com/ca2bdcbc0726ce40658353f8868be9f6

  15. superhelp101
    • one year ago
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    a sub n is -192

  16. superhelp101
    • one year ago
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    so would it be A?

  17. rational
    • one year ago
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    \[a_{150} = 1000+(150-1)(-8) = -192\]

  18. rational
    • one year ago
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    then the sum is \[S_{150} = \frac{150(1000-192)}{2} =606000 \]

  19. superhelp101
    • one year ago
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    oh ! i don't know :/

  20. superhelp101
    • one year ago
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    how would u do it?

  21. rational
    • one year ago
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    we're done!

  22. rational
    • one year ago
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    606000 is the sum of first 150 terms it seems the options are wrong

  23. superhelp101
    • one year ago
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    a) 61200 b)90000 c)122400 d)180000 these are the answer choices ?

  24. superhelp101
    • one year ago
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    it might been a typo

  25. Nnesha
    • one year ago
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    take a screenshot o^_^o

  26. superhelp101
    • one year ago
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    it ok, but thxx for all the help :D

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