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Math Help - geometric progression problem

  1. #1
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    geometric progression problem

    hello

    here's the question I need a hand with

    Find the first three terms of a geometric progression where all terms are positive, S4 (the sum to four terms) = 21 2/3 and the sum to infinity = 27

    thanks in advance, all help is much appreciated
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Wink

    Find the first three terms of a geometric progression where all terms are positive, S4 (the sum to four terms) = 21 2/3 and the sum to infinity = 27

    The formulas used for GP
    Sn=a(1-(r^n))/(1-r) for n terms...(2)

    and for infinite terms
    Si=a/(1-r) for 0<r<1.......(1)

    a=1st term
    r=constant ratio
    use it in above
    put (1) in (2) for n= 4 to get r and thus a
    further use (2) to find the terms reqd


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  3. #3
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    Hello, listeningintently!

    Find the first three terms of a geometric progression
    where all terms are positive, S_4 \:=\:\frac{65}{3}, and S_{\infty} = 27
    You're expected to know these formulas:

    . . Sum of the first n terms: . S_n \;=\;a\frac{1-r^n}{1-r}

    . . Sum to infinity: . S_{\infty} \:=\:\frac{a}{1-r}


    The sum of the first four terms is \tfrac{65}{3}\!:\quad a\frac{1-r^4}{1-r} \:=\:\frac{65}{3} .[1]

    The sum to infinity is 27: . \frac{a}{1-r} \:=\:27 \quad\Rightarrow\quad a \:=\:27(1-r) .[2]

    Substitute [2] into [1]: . 27(1-r)\cdot\frac{1-r^4}{1-r} \;=\;\frac{65}{3} \quad\Rightarrow\quad 27(1-r^4) \:=\:\frac{65}{3}

    . . 1 - r^4 \:=\:\frac{65}{81} \quad\Rightarrow\quad r^4 \:=\:\frac{16}{81}\quad\Rightarrow\quad r \:=\:\sqrt[4]{\frac{16}{81}} \quad\Rightarrow\quad \boxed{r \:=\:\tfrac{2}{3}}

    Substitute into [2]: . a \:=\:27\left(1-\tfrac{2}{3}\right) \:=\:9 \quad\Rightarrow\quad \boxed{a \:=\:9 }


    Therefore, the first three terms are: . a, \,ar,\, ar^2 \;=\;9,\:6,\:4

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