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kathert

  • 2 years 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
    • 2 years ago
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    the formula for sum of n terms is Sn = a1 * (1 - r^n) ------ 1 - r

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

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

  4. theEric
    • 2 years 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
    • 2 years ago
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    r = common ratio = second term / first term a1 = first term and n = number of terms

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

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

  8. theEric
    • 2 years ago
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    @kathert I agree with 87384 :)

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

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

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

  12. theEric
    • 2 years 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
    • 2 years ago
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    @theEric Thanks :)

  14. cwrw238
    • 2 years 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
    • 2 years ago
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    Whoa how do I do 7 root in my calculator??

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

  17. theEric
    • 2 years ago
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    No!

  18. theEric
    • 2 years ago
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    Bad me!

  19. theEric
    • 2 years ago
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    r ^ (1/7)

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

  21. kathert
    • 2 years ago
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    Haha okay gimme a minute to figure this out...

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

  23. kathert
    • 2 years ago
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    Okay okay how do i do the last two??

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

  25. theEric
    • 2 years 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
    • 2 years ago
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    So, you have \(a\) and \(r\), and you need \(n\) or \(r^n\).

  27. theEric
    • 2 years ago
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    Do you see how to get that?

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

  29. theEric
    • 2 years 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
    • 2 years ago
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    General guideline: if you want something, solve for it.

  31. kathert
    • 2 years ago
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    wait so n is -11,246??

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

  33. kathert
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Ohhhhh okay! I get it now!

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

  38. kathert
    • 2 years ago
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    I got -11 1/5

  39. theEric
    • 2 years 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
    • 2 years ago
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    okay

  41. theEric
    • 2 years 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
    • 2 years ago
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    Maybe you just had some calculator error.

  43. kathert
    • 2 years ago
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    Oh I see what I did wrong

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

  45. theEric
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    so would it be 10 (1-(781250/10))/1-r?

  49. kathert
    • 2 years ago
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    @theEric

  50. theEric
    • 2 years ago
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    Sorry! Hi!

  51. kathert
    • 2 years ago
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    Its alright I disappeared for dinner so... haha

  52. theEric
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Which you can rearrange to spell \(r\ a\ n\): fun fact..

  55. kathert
    • 2 years 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
    • 2 years ago
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    Haha, thanks! And check your "\(r^n\)", or \(5^8\).

  57. kathert
    • 2 years ago
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    so 390625?

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

  59. kathert
    • 2 years ago
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    so that goes where i put the 78125

  60. theEric
    • 2 years ago
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    Yep!

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

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

  63. kathert
    • 2 years ago
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    I got 781,248 for my answer

  64. theEric
    • 2 years ago
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    I got the same! :) Congrats!

  65. theEric
    • 2 years ago
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    Any questions about this problem?

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

  67. theEric
    • 2 years ago
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    You're welcome! Take care!

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