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Thread: 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 $\displaystyle x = \frac{1}{2}y$.


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

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


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

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

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

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


    Since $\displaystyle x = \frac{1}{2}y$ this means $\displaystyle x = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$.
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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 $\displaystyle 21\tfrac{2}{3}$ and the sum to infinity is 27.

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

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


    Divide [1] by [2]: .$\displaystyle \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} $ . $\displaystyle \Rightarrow\quad r^4 \:=\:\frac{16}{81} \quad\Rightarrow\quad\boxed{r \:=\:\frac{2}{3}}$

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


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

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    Hello again, christina!

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


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

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

    . . $\displaystyle \begin{array}{c||ccc}
    \text{Number} & A & B & C \\ \hline
    \text{Geometric}& a & ar & ar^2 \end{array}$


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

    . . $\displaystyle \begin{array}{c||ccc} \text{Number} & B & A & C \\ \hline
    \text{Arithmetic} & ar & a & ar^2 \end{array}$


    The common difference $\displaystyle d$ is the difference between consecutive terms:

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


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

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


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


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

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


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

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

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

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