Series and Sequences

**christina** · January 9th 2010, 03:32 AM

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

**Prove It** · January 9th 2010, 05:28 AM

Originally Posted by christina

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 $x = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$ .

**Soroban** · January 9th 2010, 02:10 PM

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: . $\begin{Bmatrix}a&=&9 \\ ar &=& 6 \\ ar^2 &=& 4 \end{Bmatrix}$

**Soroban** · January 9th 2010, 02:52 PM

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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