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Math Help - Sum to Infinity

  1. #1
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    Sum to Infinity

    Hi again guys!

    Once again I need help with what should be relatively simple, because once again my maths textbook refuses to explain things properly

    Ok so its about the Sum to Infinity of a Geometric Series:

    In a certain geometric sequence the first term exceeds the second term by 32 and the sum of the second and third terms is 48. Find the infinite sum of the sequence.

    I thought I had the answer, but turns out I was way off-I'll give you what I got and I would appreciate it if someone could please tell me where I went wrong. I got: a= 192/7, r=-1/6

    Because a=32/1-r and a= 48/r-r^2. Since a=a, 32/1-r = 48/r-r^2

    I solved for r by using the Quadractic formula.

    Thankssss!!
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  2. #2
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    Hello,

    Quote Originally Posted by Stevo_Evo_22 View Post
    Hi again guys!

    Once again I need help with what should be relatively simple, because once again my maths textbook refuses to explain things properly

    Ok so its about the Sum to Infinity of a Geometric Series:

    In a certain geometric sequence the first term exceeds the second term by 32 and the sum of the second and third terms is 48. Find the infinite sum of the sequence.

    I thought I had the answer, but turns out I was way off-I'll give you what I got and I would appreciate it if someone could please tell me where I went wrong. I got: a= 192/7, r=-1/6

    Because a=32/1-r and a= 48/r-r^2. Since a=a, 32/1-r = 48/r-r^2

    I solved for r by using the Quadractic formula.

    Thankssss!!
    I'm not sure about what you did....

    Let a_1 be the first term of the geometric series, and r the geometric progression.

    a_1=a_2+32
    But we also know that a_2=r \cdot a_1

    \implies \boxed{a_2=r \cdot a_2+32 \cdot r}

    We know that a_2+a_3=48, but we also know that a_3=r \cdot a_2

    \implies \boxed{a_2+r \cdot a_2=48}

    \implies a_2 \cdot (1+r)=48 \implies a_2=\frac{48}{1+r}

    Substituting this into the first boxed equation, we get :


    \frac{48}{1+r}=r \cdot \frac{48}{1+r}+32 \cdot r

    Multiply by 1+r and solve



    The infinite sum is S=a_1 \cdot \frac{1}{1-r} if |r|<1
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  3. #3
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    Whoaa, Ok now I'm very confused. How do you get "a(lower 2) = r x a(lower 2) +32r"????

    Yes, I do need it in simple terms
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  4. #4
    Moo
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    Quote Originally Posted by Moo View Post
    a_1=a_2+32
    But we also know that a_2=r \cdot a_1

    \implies \boxed{a_2=r \cdot a_2+32 \cdot r}
    You're talking about this one ?

    I substitute a_1 of the first equation in the second
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  5. #5
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    Quote Originally Posted by Moo View Post
    [snip]
    Let a_1 be the first term of the geometric series, and r the common ratio for the geometric progression.

    a_1=a_2+32
    But we also know that a_2=r \cdot a_1

    \implies \boxed{{\color{red}a_1 =r \cdot a_1 +32}} .... (1)

    We know that a_2 + a_3=48, but we also know that a_3 = r \cdot a_2

    \implies a_2 + r \cdot a_2=48

    \implies a_2 \cdot (1+r)=48

     \boxed{\implies a_2=\frac{48}{1+r}}

    But {\color{red}a_2 = r a_1}.

    Therefore {\color{red}r a_1 = \frac{48}{1+r}}

    \boxed{\color{red} \Rightarrow a_1 = \frac{48}{r(1+r)}} .... (2)

    Substituting (2) into (1), we get:

    \boxed{{\color{red}\frac{48}{r(1+r)} = \frac{48}{1+r}} +32} .... (3)

    Solve equation (3) for r. There are two solutions: r = 1/2 and r = -3. But only one of these will enable the infinite series to have a finite sum .....

    The infinite sum is S=a_1 \cdot \frac{1}{1-r} if |r|<1
    I've done some editing (mostly in red) to Moo's fine reply. Since it's a diferent way (of getting the same thing) you might follow it a bit better ....
    Last edited by mr fantastic; May 10th 2008 at 12:59 AM.
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  6. #6
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    Thankssss a million!!!! I thought you might have been a little off, Moo, but thanks anyway you did help a lot. And thanks to Mr. Fantastic, now I understand completely. The only thing is, would the whole equation work if I substituted for example a(lower 1) with just a, and a(lower 2) with ar etc.???

    Thanks again!

    Steve
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  7. #7
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    Quote Originally Posted by Stevo_Evo_22 View Post
    Thankssss a million!!!! I thought you might have been a little off, Moo, but thanks anyway you did help a lot. And thanks to Mr. Fantastic, now I understand completely. The only thing is, would the whole equation work if I substituted for example a(lower 1) with just a, and a(lower 2) with ar etc.???

    Thanks again!

    Steve
    Yes. Personally I'd work with a1, a2 etc. But if you understand it better using a, ar, ar^2 etc. then you should do it that way.
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  8. #8
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    Hi, sorry,

    I'm stuck witht he same question. I end up with 2r^2 + 5r - 6.

    I cannot factorise this for the life of me. The quadratic formula turns up weird answers, ditto for completing the square. Would I be able to get a bit of a hand with it?

    Thanks
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  9. #9
    Moo
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    I'm sorry if my writing wasn't clear...

    Quote Originally Posted by Stevo_Evo_22 View Post
    Hi, sorry,

    I'm stuck witht he same question. I end up with 2r^2 + 5r - 6.

    I cannot factorise this for the life of me. The quadratic formula turns up weird answers, ditto for completing the square. Would I be able to get a bit of a hand with it?

    Thanks
    Hm...There's a problem with your equation.

    Let's do it :

    <br />
\frac{48}{r(1+r)} = \frac{48}{1+r} +32<br />

    Multiplying both sides by r(1+r), we get :

    48=\frac{48 \cdot r \cdot (1+r)}{1+r}+32 \cdot r \cdot (1+r)

    48=48r+32(r^2+r)

    0=-48+48r+32r^2+32r

    0=-48+80r+32r^2

    Dividing by 16 :

    2r^2+5r-{\color{red}3}=0

    Is it better ?
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  10. #10
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    Quote Originally Posted by Stevo_Evo_22 View Post
    Hi, sorry,

    I'm stuck witht he same question. I end up with 2r^2 + 5r - 6.

    I cannot factorise this for the life of me. The quadratic formula turns up weird answers, ditto for completing the square. Would I be able to get a bit of a hand with it?

    Thanks
    \frac{48}{r(1+r)} = \frac{48}{1+r} +32

    \Rightarrow \frac{48}{r(1+r)} = \frac{48 r}{r(1+r)} + \frac{32r(1+r)}{r(1+r)}

    \Rightarrow 48 = 48r + 32r(1+r)

    \Rightarrow 32r^2 + 80r - 48 = 0

    \Rightarrow 2r^2 + 5r - {\color{red}3} = 0

    \Rightarrow (2r - 1)(r + 3) = 0

    etc.
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  11. #11
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    AAAAHHHHHHHHH, thanks, I knew i did something wrong!!!!!

    You've been a massive help to me, thanks a lot!
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  12. #12
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    Hello, Stevo_Evo_22!

    In a certain geometric sequence, the first term exceeds the second term by 32
    and the sum of the second and third terms is 48.
    Find the infinite sum of the sequence.
    The first three terms are: . a,\;ar,\;ar^2

    The first is 32 more than the second: . a \:=\:ar-32\quad\Rightarrow\quad a \:=\:\frac{32}{1-r}\;\;{\color{blue}[1]}

    The sum of the 2nd and 3rd is 48: . ar + ar^2 \:=\:48\quad\Rightarrow\quad ar(1+r) \:=\:48\;\;{\color{blue}[2]}

    Substitute [1] into [2]: . \frac{32}{1-r}\cdot r(1+r) \:=\:48

    . . which simplifies to: . 2r^2 + 5r - 3 \:=\:0

    . . which factors: . (2r-1)(r+3) \:=\:0

    . . and has roots: . r \;=\;\frac{1}{2},\;-3

    Since a convergent series must have |r| < 1, the ratio is: . r\:=\:\frac{1}{2}

    Substitute into [1]: . a \:=\:\frac{32}{1-\frac{1}{2}} \quad\Rightarrow\quad a\:=\:64


    Therefore, the sum is: . S \;=\;\frac{a}{1-r} \;=\;\frac{64}{1-\frac{1}{2}} \;=\;\boxed{128}

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