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Geometric series

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    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
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    (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}
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    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
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    (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.
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    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
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    (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?
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    (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.
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    (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.
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    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!
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    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.
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    (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}.
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    (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).
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    I remember this question, C2 on AQA right?
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    (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...
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    (Original post by f1mad)
    Un= a*r^(n-1)

    a= 2000 here
    What's your point? :s
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    (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.
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    (Original post by hassi94)
    What's your point? :s
    Ignore that, I realised you'd divided by 2000 .
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    (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.
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    (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.
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    (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

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