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anonymous

  • one year ago

The population of a local species of dragonfly can be found using an infinite geometric series where a1 = 42 and the common ratio is three fourths. Write the sum in sigma notation and calculate the sum that will be the upper limit of this population.

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  1. anonymous
    • one year ago
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    @jim_thompson5910

  2. jim_thompson5910
    • one year ago
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    what do you have so far?

  3. anonymous
    • one year ago
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    i have|dw:1442973187600:dw|

  4. jim_thompson5910
    • one year ago
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    you're very close

  5. anonymous
    • one year ago
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    oh and the 42(3/4)^i-1?

  6. jim_thompson5910
    • one year ago
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    it should be \[\Large \sum_{i=1}^{\infty} 42\left(\frac{3}{4}\right)^{i-1}\]

  7. anonymous
    • one year ago
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    Yes! That's what I meant. Thank-you :)

  8. jim_thompson5910
    • one year ago
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    now `calculate the sum that will be the upper limit of this population.`

  9. anonymous
    • one year ago
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    Hmmm...it wouldn't be diveregent right?

  10. anonymous
    • one year ago
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    divergent*

  11. jim_thompson5910
    • one year ago
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    you see how a = 42 and r = 3/4 right?

  12. anonymous
    • one year ago
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    yes

  13. anonymous
    • one year ago
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    ok! So we are plugging it into the geometric series formula!

  14. jim_thompson5910
    • one year ago
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    yeah since |r| < 1 is true, we can use S = a/(1-r)

  15. anonymous
    • one year ago
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    ok. So I got 168?

  16. jim_thompson5910
    • one year ago
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    me too

  17. anonymous
    • one year ago
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    Can you help me on another?

  18. jim_thompson5910
    • one year ago
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    sure

  19. anonymous
    • one year ago
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    What is the sum of the arithmetic sequence 8, 14, 20 …, if there are 24 terms?

  20. jim_thompson5910
    • one year ago
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    first term = ?? common difference = ??

  21. anonymous
    • one year ago
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    A1=8 C.D.=-6

  22. jim_thompson5910
    • one year ago
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    C.D is +6 actually

  23. jim_thompson5910
    • one year ago
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    we add 6 to each term to get the next one

  24. anonymous
    • one year ago
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    oh! Sorry :)

  25. jim_thompson5910
    • one year ago
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    now use the formula \[\Large S_n = \frac{n(2a+d(n-1))}{2}\] n = number of terms added up a = first term d = common difference

  26. anonymous
    • one year ago
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    So far, I have s24=24(16+6n-1/2)

  27. jim_thompson5910
    • one year ago
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    n = 24, so replace the n in n-1 with 24

  28. anonymous
    • one year ago
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    oh! Thanks

  29. jim_thompson5910
    • one year ago
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    \[\Large S_n = \frac{n(2a+d(n-1))}{2}\] \[\Large S_{24} = \frac{24(2*8+6(24-1))}{2}\] \[\Large S_{24} = ???\]

  30. anonymous
    • one year ago
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    Ok! Hold on lol

  31. anonymous
    • one year ago
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    1848 :)

  32. jim_thompson5910
    • one year ago
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    good

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