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Math Help - Series and Sequences

  1. #1
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    Series and Sequences

    Find x and y, given that 1,x and y are in arithmetic sequence, and 1,y,x are in geometric sequence
    Find the first three terms of a geometric sequence given that the sum of the first four terms is 21 2/3 and the sum to infinity is 27.
    The three numbers a,b and c, whose sum is 15, are successive terms of a geometric sequence, and b, a and c are successive terms of an arithmetic sequence. Find the values of a, b and c


    Thank you so much for help given on these questions
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  2. #2
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    Quote Originally Posted by christina View Post
    Find x and y, given that 1,x and y are in arithmetic sequence, and 1,y,x are in geometric sequence
    Find the first three terms of a geometric sequence given that the sum of the first four terms is 21 2/3 and the sum to infinity is 27.
    The three numbers a,b and c, whose sum is 15, are successive terms of a geometric sequence, and b, a and c are successive terms of an arithmetic sequence. Find the values of a, b and c


    Thank you so much for help given on these questions
    If you have an arithmetic sequence 1, x, y, ... then the common difference is x.

    So y = x + x = 2x, or x = \frac{1}{2}y.


    If you have a geometric sequence 1, y, x, ... then the common ratio is y.

    So x = y\cdot y = y^2.


    Therefore y^2 = \frac{1}{2}y

    y^2 - \frac{1}{2}y = 0

    y\left(y - \frac{1}{2}\right)= 0

    y = 0 or y = \frac{1}{2}.


    Since x = \frac{1}{2}y this means x = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}.
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  3. #3
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    Hello, christina!

    Here's the second one . . .


    Find the first three terms of a geometric sequence,
    given that the sum of the first four terms is 21\tfrac{2}{3} and the sum to infinity is 27.

    \text{The sum of the first four terms is }21\tfrac{2}{3}:\;\;a\,\frac{1-r^4}{1-r} \:=\:\frac{65}{3} .[1]

    \text{The sum to infinity is 27: }\;\;\frac{a}{1-r} \:=\:27 .[2]


    Divide [1] by [2]: . \frac{\;a\,\dfrac{1-r^4}{1-r}\;} {\dfrac{a}{1-r}} \;=\;\frac{\;\dfrac{65}{3}\;} {27} \quad\Rightarrow\quad 1-r^4 \:=\:\frac{65}{81} . \Rightarrow\quad r^4 \:=\:\frac{16}{81} \quad\Rightarrow\quad\boxed{r \:=\:\frac{2}{3}}

    Substitute into [2]: . \frac{a}{1-\frac{2}{3}} \:=\:27 \quad\Rightarrow\quad\boxed{ a \:=\:9}


    Therefore, the first three terms are: . \begin{Bmatrix}a&=&9 \\ ar &=& 6 \\ ar^2 &=& 4 \end{Bmatrix}

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  4. #4
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    Hello again, christina!

    And here's the last one . . . and there are three solutions.


    The three numbers A,B,C, whose sum is 15,
    are successive terms of a geometric sequence,
    and B,A,C are successive terms of an arithmetic sequence.
    Find the values of A,B,\text{ and }C.

    In the geometric series, the numbers are represented like this:

    . . \begin{array}{c||ccc}<br />
\text{Number} & A & B & C \\ \hline<br />
\text{Geometric}& a & ar & ar^2 \end{array}


    In the arithmetic series, the same numbers are represented like this:

    . . \begin{array}{c||ccc} \text{Number} & B & A & C \\ \hline<br />
\text{Arithmetic} & ar & a & ar^2 \end{array}


    The common difference d is the difference between consecutive terms:

    . . . . \begin{array}{cccc}d &=& a - ar & {\color{blue}[1]} \\ d &=& ar^2 - a & {\color{blue}[2]} \end{array}


    Equate {\color{blue}[1]} and {\color{blue}[2]}: . a-ar \:=\:ar^2-a \qquad\Rightarrow\qquad ar^2 + ar - 2a \:=\:0

    . . a(r-1)(r+2) \:=\:0 \qquad\Rightarrow\qquad a = 0\;\text{ or }\;r = 1\;\text{ or }\;r= -2


    If a = 0, we have a trivial sequence: . \boxed{\begin{array}{cc}\text{Geometric:} & 0,0,0 \\ \text{Arithmetic:}& 0,0,0\end{array}}


    We have: . A + B + C \:=\:15 .[3]

    If r = 1, we have another trivial sequence: . \boxed{\begin{array}{cc}\text{Geometric:} & 5,5,5 \\ <br />
\text{Arithmetic:} & 5,5,5 \end{array}}


    [3] becomes: . a + ar + ar^2 \:=\:15 \quad\Rightarrow\quad a(1+r+r^2) \:=\:15

    If r = \text{-}2, we have: . a(4-2+1) \:=\:15 \quad\Rightarrow\quad a \:=\:3

    And we have a third solution: . \boxed{\begin{array}{cccc}\text{Geometric:} & 3& \text{-}6 & 12 \\ \text{Arithmetic:} & \text{-}6 & 3 & 12 \end{array}}

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