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Math Help - Geometric sequence problem

  1. #1
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    Geometric sequence problem

    "Given a_{n} = 5, r = \frac{1}{2}, and S_{n} = 155 for a geometric sequence, find a_{1} and n."

    Using the formulas a_{n} = a_{1}(r)^{n-1} and S_{n} = \frac{a_{1} (1- r^{n})}{1 - r}, I start with these two equations:

    5 = a_{1}\frac{1}{2}^{n-1} and

    155 = \frac{a_{1}(1 - \frac{1}{2}^{n})}{1 - \frac{1}{2}}

    Simplifying each for a_{1}, I get

    a_{1} = \frac{5}{\frac{1}{2}^{n-1}} and

    a_{1} = \frac{155}{2(1 - \frac{1}{2}^{n})}

    Therefore,

    \frac{5}{\frac{1}{2}^{n-1}} = \frac{155}{2(1 - \frac{1}{2}^{n})}

    I'm stuck at this point... how do I finish simplifying this equation? I'm guessing it'll involve logs... Any help would be greatly appreciated.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by earachefl View Post
    "Given a_{n} = 5, r = \frac{1}{2}, and S_{n} = 155 for a geometric sequence, find a_{1} and n."

    Using the formulas a_{n} = a_{1}(r)^{n-1} and S_{n} = \frac{a_{1} (1- r^{n})}{1 - r}, I start with these two equations:

    5 = a_{1}\frac{1}{2}^{n-1} and

    155 = \frac{a_{1}(1 - \frac{1}{2}^{n})}{1 - \frac{1}{2}}

    Simplifying each for a_{1}, I get

    a_{1} = \frac{5}{\frac{1}{2}^{n-1}} and

    a_{1} = \frac{155}{2(1 - \frac{1}{2}^{n})}

    Therefore,

    \frac{5}{\frac{1}{2}^{n-1}} = \frac{155}{2(1 - \frac{1}{2}^{n})}

    I'm stuck at this point... how do I finish simplifying this equation? I'm guessing it'll involve logs... Any help would be greatly appreciated.
    no logarithms needed here, just some "simple" simplification

    you seem to be on the money so far. i will restart the problem though. you way is fine, but it seems having the halves around was confusing you. i will rewrite the problem using powers of 2, not 1/2

    a_n = 5, r = \frac 12, \mbox{ and } S_n = 155

    a_n = ar^{n - 1}

    \Rightarrow 5 = a \left( \frac 12 \right)^{n - 1} = a \cdot 2^{1 - n}

    \Rightarrow a = 5 \cdot 2^{n - 1} ....................(1)


    S_n = \frac {a \left( 1 - r^n \right)}{1 - r}

    \Rightarrow 155 = \frac {a \left( 1 - 2^{-n} \right)}{\frac 12}

    \Rightarrow \frac {155}2 = a \left( 1 - 2^{-n} \right)

    \Rightarrow a = \frac {155}{2 - 2^{1 - n}} .....................(2)

    equating (1) and (2), we obtain:

    5 \cdot 2^{n - 1} = \frac {155}{2 - 2^{1 - n}}

    \Rightarrow 2^{n - 1} = \frac {31}{2 - 2^{1 - n}}

    \Rightarrow 2^n - 1 = 31

    \Rightarrow 2^n = 32

    \Rightarrow \boxed{n = 5}


    back-substitute into (1):

    \Rightarrow a = 5 \cdot 2^4

    \Rightarrow \boxed{a = 80}


    if you insist on picking up where you left off:

    ... \frac{5}{\frac{1}{2}^{n-1}} = \frac{155}{2(1 - \frac{1}{2}^{n})}
    \Rightarrow \frac 5{2^{1 - n}} = \frac {155}{2 \left( 1 - 2^{-n} \right)} .........divide both sides by 5

    \Rightarrow \frac 1{2^{1 - n}} = \frac {31}{2 \left( 1 - 2^{-n} \right)} ............multiply both sides by 2 \left( 1 - 2^{-n} \right)

    \Rightarrow \frac {2 \left( 1 - 2^{-n} \right)}{2^{1 - n}} = 31

    \Rightarrow 2^{1 - (1 - n)} \left( 1 - 2^{-n} \right) = 31

    \Rightarrow 2^n \left( 1 - 2^{-n} \right) = 31

    \Rightarrow 2^n - 1 = 31

    and the rest is trivial
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  3. #3
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    i will rewrite the problem using powers of 2, not 1/2

    <br />
\Rightarrow 5 = a \left( \frac 12 \right)^{n - 1} = a \cdot 2^{1 - n}<br />


    So when you are doing this, effectively, you are doing

    a \frac 12 ^{(-1)(n - 1)(-1)}? Is that the trick?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by earachefl View Post
    So when you are doing this, effectively, you are doing

    a \frac 12 ^{(-1)(n - 1)(-1)}? Is that the trick?
    no.
    \frac 12 = 2^{-1}

    so the trick is: a \left( \frac 12 \right)^{n - 1} = a \left( 2^{-1} \right)^{n - 1} = a \cdot 2^{(-1)(n -1)} = a \cdot 2^{1 - n}
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  5. #5
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    Hello, earachefl!

    Another approach . . .


    Given: . a_n \,=\, 5,\;\;r \,=\, \frac{1}{2},\;\;S_n \,=\, 155 . for a geometric sequence.
    Find: a_1 and n

    We have: . a_n \;=\;a_1\left(\frac{1}{2}\right)^{n-1}\quad\Rightarrow\quad \frac{a_1}{2^{n-1}}\;=\;5 .[1]

    and: . S_n \;=\;a_1\,\frac{1 - \left(\frac{1}{2}\right)^n}{1-\frac{1}{2}} \;=\;a_1\,\frac{1-\frac{1}{2^n}}{\frac{1}{2}} \;=\;2a_1\left(1-\frac{1}{2^n}\right)\quad\Rightarrow\quad 2a_1 \,- \,\frac{a_1}{2^{n-1}} \;=\;155 .[2]


    Add [1] and [2]: . 2a_1 \:=\:160\quad\Rightarrow\quad\boxed{ a_1 \:=\:80}


    Substitute into [1]: . \frac{80}{2^{n-1}} \:=\:5\quad\Rightarrow\quad 2^{n-1} \:=\:16\quad\Rightarrow\quad 2^{n-1} \:=\:2^4

    Therefore: . n-1\:=\:4\quad\Rightarrow\quad\boxed{n \:=\:5}

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