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kathert

  • one year ago

Please help... Only 2 days left for me on a summer school class and I am lost... 1. What is the sum of the geometric sequence 8, -16, 32 . if there are 15 terms? (1 point) 2. What is the sum of the geometric sequence 4, 12, 36 . if there are 9 terms? (1 point) 3. What is the sum of a 6-term geometric sequence if the first term is 11, the last term is -11,264 and the common ratio is -4? (1 point) 4. What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is 781,250? (1 point) Show all work as well

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  1. cwrw238
    • one year ago
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    the formula for sum of n terms is Sn = a1 * (1 - r^n) ------ 1 - r

  2. theEric
    • one year ago
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    http://www.mathsisfun.com/algebra/sequences-sums-geometric.html

  3. kathert
    • one year ago
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    But what do I plug in there to get the answers?

  4. theEric
    • one year ago
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    The link I posted will walk you through what geometric sequences are, and that formula to get the answers!

  5. cwrw238
    • one year ago
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    r = common ratio = second term / first term a1 = first term and n = number of terms

  6. cwrw238
    • one year ago
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    so for question 1 r = -16/8 = -2 a1 = 8 and n = 15

  7. kathert
    • one year ago
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    So would the answer for the first one be 87384?

  8. theEric
    • one year ago
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    @kathert I agree with 87384 :)

  9. cwrw238
    • one year ago
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    8 * ( 1 - (-2)^15) ------------- = 87384 1 - (-2)

  10. kathert
    • one year ago
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    Oh yay! I got it right! Sorry I suck at math...

  11. cwrw238
    • one year ago
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    A GS can be written as a, ar , ar^2 , ar^3 .......

  12. theEric
    • one year ago
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    I've \(always\) been slow at math. But it took practice, and getting help, and I can do more math things now! Best of luck in your class! :) And it looks like you're in good hands with cwrw238 .

  13. kathert
    • one year ago
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    @theEric Thanks :)

  14. cwrw238
    • one year ago
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    for the last probem you can find the common ratio r by dividing the 8th term by the first then you take the 7th root 8th term = ar^7 8th term / first term = ar^7 / a = r^7 781,250 / 10 = 78125 now use your calculator to find the 7th root of 78125 then use the sum formula gotta go now

  15. kathert
    • one year ago
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    Whoa how do I do 7 root in my calculator??

  16. theEric
    • one year ago
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    Do it like this: r ^ (- 7)

  17. theEric
    • one year ago
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    No!

  18. theEric
    • one year ago
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    Bad me!

  19. theEric
    • one year ago
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    r ^ (1/7)

  20. theEric
    • one year ago
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    That's what you want, \[\Huge r^{\frac{1}{7}}\]

  21. kathert
    • one year ago
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    Haha okay gimme a minute to figure this out...

  22. kathert
    • one year ago
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    I got 39364 for number 2 is that right?

  23. kathert
    • one year ago
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    Okay okay how do i do the last two??

  24. theEric
    • one year ago
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    That's what I got for #2 as well.

  25. theEric
    • one year ago
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    3. What is the sum of a 6-term geometric sequence if the first term is 11, the last term is -11,264 and the common ratio is -4? (1 point) Well, the first term is your \(a\).\[a=11\]The last term is your \(a\ r^n\).\[a\ r^n=-11,264\]The common ratio is your \(r\).\[r=-4\] You need \(\Large a\frac{1-r^n}{1-r}\).

  26. theEric
    • one year ago
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    So, you have \(a\) and \(r\), and you need \(n\) or \(r^n\).

  27. theEric
    • one year ago
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    Do you see how to get that?

  28. kathert
    • one year ago
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    uhhh.... no not really...so it would be -11 (1-(-4^n)/(1--4)?

  29. theEric
    • one year ago
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    Yep! Hey, you know \(a\ r^n=-11,264\), so you can solve for \(r^n\)! That's how you'll finish that problem.

  30. theEric
    • one year ago
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    General guideline: if you want something, solve for it.

  31. kathert
    • one year ago
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    wait so n is -11,246??

  32. theEric
    • one year ago
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    No, \(a\ r^n=-11,264\). So you divide both sides by \(a=11\). That's algebra!

  33. kathert
    • one year ago
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    im sorry but im so lost.... where does the -11246 come in? do i set it equal to the equation?

  34. theEric
    • one year ago
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    It is necessary to find the \(r^n\). Let me show you. Are you familiar with algebra? \[a\ r^n = -11,264\]and\[a=11\]By substituting \(11\) for \(a\), which is okay because it's the same value either way, you'll get:\[11\ r^n=-11,264\]Now, you want to get \(r^n\) alone. So what you do is, you divide both sides by \(11\). 1. If the two sides are equal, and you do the same thing to both sides, both sides will still be equal! 2. Why divide by \(11\)? Well \(r^n\) is being multiplied by \(11\), and so you want to negate that. You want to make it be \(\Large \frac{\cancel{11}\ r^n}{\cancel{11}}\) So, we left our equation off at\[11\ r^n=-11,264\]We divide by \(11\) to get\[\frac{11\ r^n}{11}=\frac{-11,264}{11}\]\[\frac{\cancel{11}\ r^n}{\cancel{11}}=\frac{-11,264}{11}\]\[r^n=\frac{-11,264}{11}\]

  35. theEric
    • one year ago
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    Since you now have \(r^n\), \(a\), and \(r\), you can use that formula that you used for #1 and #2.

  36. kathert
    • one year ago
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    Ohhhhh okay! I get it now!

  37. theEric
    • one year ago
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    SWEET! :) So, we'll both calculate #3 and see what we get....

  38. kathert
    • one year ago
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    I got -11 1/5

  39. theEric
    • one year ago
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    I got \(2255\)... Let me use Wolfram Alpha to double check. Then I can show you a link to the math.

  40. kathert
    • one year ago
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    okay

  41. theEric
    • one year ago
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    http://www.wolframalpha.com/input/?i=11*%281-%28-11264%2F11%29%29%2F%281-%28-4%29%29

  42. theEric
    • one year ago
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    Maybe you just had some calculator error.

  43. kathert
    • one year ago
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    Oh I see what I did wrong

  44. kathert
    • one year ago
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    How would I go about starting the last one?

  45. theEric
    • one year ago
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    Well, I'm sure you know the formula you need to use, by now!\[\text{sum}=a\frac{(1-r^n)}{(1-r)}\] 4. What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is 781,250? (1 point) You need \(a\), \(r\), and \(r^n\), or \(n\). What do you know from the problem, about the geometric sequence?

  46. theEric
    • one year ago
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    Refresher: \(a\) is the first term, or the common multiplier. \(r\) is the common ratio. \(n\) is the number of terms in the sequence.

  47. theEric
    • one year ago
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    8-term \(\rightarrow n=8\) first term is 10 \(\rightarrow a=10\) last term is 781,250 \(\rightarrow a\ r^{n-1} =781,250\)

  48. kathert
    • one year ago
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    so would it be 10 (1-(781250/10))/1-r?

  49. kathert
    • one year ago
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    @theEric

  50. theEric
    • one year ago
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    Sorry! Hi!

  51. kathert
    • one year ago
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    Its alright I disappeared for dinner so... haha

  52. theEric
    • one year ago
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    Nope, sorry! Small mistake! \[a\ r^{n-1} =781,250\]\(\qquad\Downarrow\qquad\)Substitute \(8\) for \(n\) \[a\ r^{8-1} =a\ r^{7}=781,250\]\(\qquad\Downarrow\qquad\)Divide both sides by \(a\), and then substitute \(10\) in for \(a\) \[r^7=\frac{781,250}{a}=\frac{781,250}{10}=78,125\]\(\qquad\Downarrow\qquad\)Get the seventh root of both sides\[\sqrt[7]{r^7}=r=\sqrt[7]{78,125}=5\]

  53. theEric
    • one year ago
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    Well, you knew you didn't know \(r\), so I guess your only mistake was substituting \(r^n\) with \(r^{n-1}=781,250\), but you definitely had the right idea otherwise! I just found \(r\) for you, then... Any questions on that part? Now you have \(a\), \(n\), and \(r\).

  54. theEric
    • one year ago
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    Which you can rearrange to spell \(r\ a\ n\): fun fact..

  55. kathert
    • one year ago
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    so it would be 8 *(1-78125)/(1-5) just to be sure nice fun fact btw haha

  56. theEric
    • one year ago
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    Haha, thanks! And check your "\(r^n\)", or \(5^8\).

  57. kathert
    • one year ago
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    so 390625?

  58. theEric
    • one year ago
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    You got \(78125\) from \(r^{n-1}\), so I see where that came from :) And, yep! \(390625\).

  59. kathert
    • one year ago
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    so that goes where i put the 78125

  60. theEric
    • one year ago
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    Yep!

  61. theEric
    • one year ago
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    It is \(r^n\), after all.

  62. theEric
    • one year ago
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    and your formula is\[\text{sum}=a\frac{(1-r^n)}{(1-r)}\]

  63. kathert
    • one year ago
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    I got 781,248 for my answer

  64. theEric
    • one year ago
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    I got the same! :) Congrats!

  65. theEric
    • one year ago
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    Any questions about this problem?

  66. kathert
    • one year ago
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    Nope I'm good! Thank you for your help!

  67. theEric
    • one year ago
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    You're welcome! Take care!

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