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anonymous

  • one year ago

Find S11 for 3 + 1 + (-1) + (-3) +… A. -77 B. -75 C. -73 D. -71

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  1. jdoe0001
    • one year ago
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    hmm doesn't look like a sequence to me for one

  2. jdoe0001
    • one year ago
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    can you post a quick screenshot of the material? so we can see what you mean

  3. Astrophysics
    • one year ago
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    Nope, it's a series

  4. jdoe0001
    • one year ago
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    hmm ok

  5. Astrophysics
    • one year ago
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    So we can use sum of an arithmetic series equation I guess which is \[S_n = \frac{ n }{ 2 }[2a+(n-a)d]\] a = first term in series, d = common difference

  6. Astrophysics
    • one year ago
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    n is what you're looking for in this case n = 11

  7. Astrophysics
    • one year ago
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    So what's the common difference, look at your previous post what I wrote when you left

  8. Astrophysics
    • one year ago
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    So I'll post it again for reference, the common difference is \[t_2-t_1=t_3-t_2\]

  9. anonymous
    • one year ago
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    I'm really confused right now..

  10. Astrophysics
    • one year ago
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    I would say so you jumped from a sequence from your last question to a series here, so just focus on the equation for now. The t here represents the term and the subscript is the address of the term, so \[t_1 = 3~~~t_2 = 1\] etc

  11. welshfella
    • one year ago
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    Common difference can be worked out by subtracting the first term from the second

  12. Astrophysics
    • one year ago
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    \[t_2-t_1 = t_3-t_2 \implies 1-3 = -1-1 \implies -2 = -2 \checkmark\] so it works out, our common difference d = -2

  13. Astrophysics
    • one year ago
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    With me so far?

  14. anonymous
    • one year ago
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    yes!

  15. anonymous
    • one year ago
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    i'm just writing everything down in my notebook!

  16. Astrophysics
    • one year ago
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    Ok cool :) Well now lets find a, what is a?

  17. anonymous
    • one year ago
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    is already one of the terms given, or do I have to find it?

  18. Astrophysics
    • one year ago
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    a = first term in series

  19. anonymous
    • one year ago
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    A not already sorry

  20. anonymous
    • one year ago
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    so its 3

  21. Astrophysics
    • one year ago
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    Good!

  22. Astrophysics
    • one year ago
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    So we now have d = -2, a = 3, n = 11

  23. Astrophysics
    • one year ago
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    Now we can use the equation I mentioned above \[S_n = \frac{ n }{ 2 }[2a+(n-a)d]\] can you plug it all in now

  24. anonymous
    • one year ago
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    sn=11/2 (2(3)+(11+3)-2) correct?

  25. Astrophysics
    • one year ago
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    Close, your sign in (n-a) is off but it looks good \[S_{11} = \frac{ 11 }{ 2 }[2(3)+(11-3)(-2)]\]

  26. anonymous
    • one year ago
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    i got -55 ?

  27. Astrophysics
    • one year ago
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    Yup, that looks good

  28. anonymous
    • one year ago
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    but it isn't one of the choices?

  29. Astrophysics
    • one year ago
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    Mhm...

  30. Astrophysics
    • one year ago
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    So is your question asking for \[S_{11}\]

  31. anonymous
    • one year ago
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    yes!

  32. Astrophysics
    • one year ago
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    So it's exactly Find S11 for 3 + 1 + (-1) + (-3) +… right

  33. anonymous
    • one year ago
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    yes!

  34. Astrophysics
    • one year ago
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    Mhm, everything looks right

  35. Astrophysics
    • one year ago
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    Ok lets see...a = 3, d = -2, n = 11

  36. Astrophysics
    • one year ago
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    I have located the problem it's the original equation I gave you, it should (n-1) not (n-a) sorry! \[S_n = \frac{ n }{ 2 }[2a+(n-1)d] = \frac{ 11 }{ 2 }[2(3)+(11-1)(-2)] = \]

  37. Astrophysics
    • one year ago
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    Try now :)

  38. Astrophysics
    • one year ago
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    So the equation is \[S_n = \frac{ n }{ 2 }[2a+(n-\color{red}1)d]\]

  39. anonymous
    • one year ago
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    I got 72.8? so -73?

  40. Astrophysics
    • one year ago
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    Nope, you must've put in wrong

  41. Astrophysics
    • one year ago
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    \[\frac{ 11 }{ 2 }[2(3)+(11-1)(-2)] = \frac{ 11 }{ 2 }[6+10(-2)] = \frac{ 11 }{ 2 }[6-20]=\frac{ 11 }{ 2 }(-14) = -77\]

  42. Astrophysics
    • one year ago
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    Use your order of operations :-)

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