Geometric series

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  1. gavinlee's Avatar
    1 02-05-2012  07:33
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    Geometric series
    Hi everyone,

    I've got 3 geometric series questions that I'm struggling with. It would be great if anyone could give me a few pointers.

    1)A geometric series has has 4th term equal to 8 times the 7th term. First term is 1024. Find the sum of the first 15 terms.

    OK, so as I know the first term, I need the common ratio. I started with ar^3=8ar^6 but couldn't go on from there as I don't know any of the values except for a=1024.

    2)The sum of the first 4 terms of a geometric series is 5468.75 and a=2000. Find the value of the common ratio if all the terms are positive.

    OK, so this is what I did:
    \frac{5468.75}{2000}=r^3=2.734.. .\Rightarrow r=\sqrt[3]{2.734}=1.398 however it's not the right answer.

    3)The sum of the first n terms of the geometric series is: 3-6+12-... is 129. Find n.

    So:
    129_n = \frac{3(1-(-2^n))}{1-(-2)} but I really don't know how to derive n when I don't know r!

    I'm stuck
  2. raheem94's Avatar
    2 02-05-2012  08:03
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    • Posts: 5,512
    Re: Geometric series
    (Original post by gavinlee)
    Hi everyone,

    I've got 3 geometric series questions that I'm struggling with. It would be great if anyone could give me a few pointers.

    1)A geometric series has has 4th term equal to 8 times the 7th term. First term is 1024. Find the sum of the first 15 terms.

    OK, so as I know the first term, I need the common ratio. I started with ar^3=8ar^6 but couldn't go on from there as I don't know any of the values except for a=1024.
    Question 1,

    ar^3=8ar^6

    Now divide both sides by 'a' to get, r^3=8r^6 \implies 8r^6-r^3 = 0
    Find 'r' from it.

    For the sum, just use the formula, \displaystyle \frac{a(1-r^n)}{1-r}
  3. bencrossley1's Avatar
    3 02-05-2012  08:10
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    Re: Geometric series
    For Q1

    only using the equation a r^3 = 8 a r^6 you should notice that you have 1 equation and 1 unknown, (silent hooray) this is good because this means you are guaranteed a solution.

    Dividing through by a and r^3 leaves you with the equation 1= 8 r^3 which you can then solve to find r = 0.5

    Now input your known values into the formula, and you should get an answer of S_n = 2047.9375


    For Q3

    r = -2, a = 3

    hence using S_n = a\frac{1-r^n}{1-r} we get:

    129 = 3\frac{1-(-2)^n}{1-(-2)}

    129 = 3\frac{1-(-2)^n}{3}

    129 = 1-(-2)^n

    128 = -(-2)^n

    -128 = (-2)^n \Rightarrow n=7



    I need to go to uni, so hopefully someone will help with Q2 or maybe this will be enough for you to work out how to do it. Good luck
  4. raheem94's Avatar
    4 02-05-2012  08:16
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    Re: Geometric series
    (Original post by bencrossley1)
    For Q1

    only using the equation a r^3 = 8 a r^6 you should notice that you have 1 equation and 1 unknown, (silent hooray) this is good because this means you are guaranteed a solution.

    Dividing through by a and r^3 leaves you with the equation 1= 8 r^3 which you can then solve to find r = 0.5

    Now input your known values into the formula, and you should get an answer of S_n = 2047.9375


    For Q3

    r = -2, a = 3

    hence using S_n = a\frac{1-r^n}{1-r} we get:

    129 = 3\frac{1-(-2)^n}{1-(-2)}

    129 = 3\frac{1-(-2)^n}{3}

    129 = 1-(-2)^n

    128 = -(-2)^n

    -128 = (-2)^n \Rightarrow n=7



    I need to go to uni, so hopefully someone will help with Q2 or maybe this will be enough for you to work out how to do it. Good luck
    Don't mind but please avoid giving full solutions. You are not very active member of TSR so might not be aware of the rules, full solutions are considered a last resort.
    Post rating:
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  5. gavinlee's Avatar
    5 02-05-2012  09:48
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    Re: Geometric series
    Thanks Raheem and Ben.


    Does anyone know why this is wrong:


    \frac{5468.75}{2000}=r^3=2.734.. .\Rightarrow r=\sqrt[3]{2.734}=1.398

    For this question:

    The sum of the first 4 terms of a geometric series is 5468.75 and a=2000. Find the value of the common ratio if all the terms are positiv
  6. raheem94's Avatar
    6 02-05-2012  09:56
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    Re: Geometric series
    (Original post by gavinlee)
    Thanks Raheem and Ben.


    Does anyone know why this is wrong:


    \frac{5468.75}{2000}=r^3=2.734.. .\Rightarrow r=\sqrt[3]{2.734}=1.398

    For this question:

    The sum of the first 4 terms of a geometric series is 5468.75 and a=2000. Find the value of the common ratio if all the terms are positiv
    For this question you have to use the formula,
    \displaystyle S_n=\frac{a(1-r^n)}{1-r}

    \displaystyle S_4 = \frac{2000(1-r^4)}{1-r} =5468.75 \implies r^4-2.734375r+1.734375 =0

    I don't know why we are getting such a complex equation.

    What is the answer of this question?
  7. Intriguing Alias's Avatar
    7 02-05-2012  10:26
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    • Location: Yorkshire
    Re: Geometric series
    (Original post by raheem94)
    For this question you have to use the formula,
    \displaystyle S_n=\frac{a(1-r^n)}{1-r}

    \displaystyle S_4 = \frac{2000(1-r^4)}{1-r} =5468.75 \implies r^4-2.734375r+1.734375 =0

    I don't know why we are getting such a complex equation.

    What is the answer of this question?
    Does seem like a very awkward question.

    One way (though this does seem a bit wishy-washy) is to say 1 + r + r^2 + r^3 = 5468.75/2000 = 175/64 (now this is the bit I find very unreasonable)

    then (1+r^2)(1+r)=175/64

    (16+16r^2)(4+4r)=175

    then r = 3/4 works by inspection.

    VERY dodgy though and I definitely think this is a pretty ridiculous question unless it was meant to be solved by iterative methods.
  8. raheem94's Avatar
    8 02-05-2012  10:33
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    Re: Geometric series
    (Original post by hassi94)
    Does seem like a very awkward question.

    One way (though this does seem a bit wishy-washy) is to say 1 + r + r^2 + r^3 = 5468.75/2000 = 175/64 (now this is the bit I find very unreasonable)

    then (1+r^2)(1+r)=175/64

    (16+16r^2)(4+4r)=175

    then r = 3/4 works by inspection.

    VERY dodgy though and I definitely think this is a pretty ridiculous question unless it was meant to be solved by iterative methods.
    I actually have tried the way you are saying, by adding the first 4 terms. But i think that some value might be wrong in the question, or there might be something missing here. Hence, i left it, i don't see any reason for a GP question to be so awkward.

    By the way, congrats for winning the member of the month award.
  9. Intriguing Alias's Avatar
    9 02-05-2012  11:00
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    Re: Geometric series
    Well I mean its clear that r is greater than 1/2 and smaller than 1 so solving it iteratively probably wouldn't be a massive pain but yes it does seem very odd for a GP question


    And thank you!
  10. gavinlee's Avatar
    10 02-05-2012  11:33
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    Re: Geometric series
    I'm glad I'm not the only one who thinks it's awkward! I haven't got the answer on me at the moment but when I get back home I'll post it.

    Thanks very much for trying to help me.
    Gav.
  11. gavinlee's Avatar
    11 02-05-2012  17:27
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    Re: Geometric series
    (Original post by gavinlee)
    Thanks Raheem and Ben.


    Does anyone know why this is wrong:


    \frac{5468.75}{2000}=r^3=2.734.. .\Rightarrow r=\sqrt[3]{2.734}=1.398

    For this question:

    The sum of the first 4 terms of a geometric series is 5468.75 and a=2000. Find the value of the common ratio if all the terms are positiv
    The answer to this is \frac{3}{4}.
  12. f1mad's Avatar
    12 02-05-2012  17:30
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    Re: Geometric series
    (Original post by gavinlee)
    The answer to this is \frac{3}{4}.
    a+ ar+ar^2+ar^3=5468.75

    2000+ 2000r+ 2000r^2+ 2000r^3= 5468.75

    r^3+ r^2+ r- 111/64=0

    (cheat here... you know r=3/4 is a factor).
    Last edited by f1mad; 02-05-2012 at 17:33.
  13. f1mad's Avatar
    13 02-05-2012  17:32
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    Re: Geometric series
    I remember this question, C2 on AQA right?
  14. f1mad's Avatar
    14 02-05-2012  17:36
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    Re: Geometric series
    (Original post by raheem94)
    For this question you have to use the formula,
    \displaystyle S_n=\frac{a(1-r^n)}{1-r}

    \displaystyle S_4 = \frac{2000(1-r^4)}{1-r} =5468.75 \implies r^4-2.734375r+1.734375 =0

    I don't know why we are getting such a complex equation.

    What is the answer of this question?
    I get a pretty reasonable answer...
  15. Intriguing Alias's Avatar
    15 02-05-2012  17:38
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    Re: Geometric series
    (Original post by f1mad)
    Un= a*r^(n-1)

    a= 2000 here
    What's your point? :s
  16. f1mad's Avatar
    16 02-05-2012  17:39
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    Re: Geometric series
    (Original post by hassi94)

    One way (though this does seem a bit wishy-washy) is to say 1 + r + r^2 + r^3 = 5468.75/2000 = 175/64 (now this is the bit I find very unreasonable)

    then (1+r^2)(1+r)=175/64

    (16+16r^2)(4+4r)=175

    then r = 3/4 works by inspection.

    VERY dodgy though and I definitely think this is a pretty ridiculous question unless it was meant to be solved by iterative methods.
    No, it factorises as a cubic .

    Subtract 5468.75/2000 from both sides.
  17. f1mad's Avatar
    17 02-05-2012  17:40
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    Re: Geometric series
    (Original post by hassi94)
    What's your point? :s
    Ignore that, I realised you'd divided by 2000 .
  18. Intriguing Alias's Avatar
    18 02-05-2012  17:42
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    Re: Geometric series
    (Original post by f1mad)
    No, it factorises as a cubic .

    Subtract 5468.75/2000 from both sides.
    But how does one just know 3/4 is a factor as you said.
  19. f1mad's Avatar
    19 02-05-2012  17:44
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    Re: Geometric series
    (Original post by hassi94)
    But how does one just know 3/4 is a factor as you said.
    You can cheat I guess .

    Or you can find possible factors of 111/64

    +-(1,3,37,111)/ +-(1,2,4,8,16,32,64).

    Long process, but it works.
  20. Intriguing Alias's Avatar
    20 02-05-2012  17:53
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    • Location: Yorkshire
    Re: Geometric series
    (Original post by f1mad)
    You can cheat I guess .

    Or you can find possible factors of 111/64

    +-(1,3,37,111)/ +-(1,2,4,8,16,32,64).

    Long process, but it works.
    In the end you're still doing what I'm doing, by inspection, which I feel is a bit silly...

    Didn't mean silly to do, silly of aqa to require it
    Last edited by Intriguing Alias; 02-05-2012 at 17:55.
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