1. ## [SOLVED] Sequences (geometric)

x, y and z are consecutive terms in a geometric sequence. If
x+y+z =7/3, and
x^2 + y^2 + z^2 = 91/9, find the values of x, y and z.

2. Originally Posted by HelenaStage
x, y and z are consecutive terms in a geometric sequence. If
x+y+z =7/3, and
x^2 + y^2 + z^2 = 91/9, find the values of x, y and z.
The general form of a geometric series is $\sum_{k=1}^n ar^k$

So $x=x$, and y in the next term so $y=rx$ and $z=ry=r^2x$

So $x+rx+r^2x=\frac{7}{3}=x(1+r+r^2)$

And the second equation tells us

$
x^2+(rx)^2+(r^2x)^2=\frac{91}{9}=x^2+r^2x^2+r^4x^2 =x^2(1+r^2+r^4)
$

So you just have to solve those equations

3. Originally Posted by artvandalay11
The general form of a geometric series is $\sum_{k=1}^n ar^k$

So $x=x$, and y in the next term so $y=rx$ and $z=ry=r^2x$

So $x+rx+r^2x=\frac{7}{3}=x(1+r+r^2)$

And the second equation tells us

$
x^2+(rx)^2+(r^2x)^2=\frac{91}{9}=x^2+r^2x^2+r^4x^2 =x^2(1+r^2+r^4)
$

So you just have to solve those equations
How? By substitution I get a quite complicated equation with radicals...

4. it seems to me that the best way to go about it is substitute for x, and yes it is going to get messy

using an equation solver it looks like x=1/3 and r=-3 or x=3 and r=-1/3

so those are the solutions you should get

5. Originally Posted by HelenaStage
How? By substitution I get a quite complicated equation with radicals...
HI

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$x(1+r+r^2)=\frac{7}{3}$ --- 1

$x^2(1+r^2+r^4)=\frac{91}{9}$ --- 2

From 1

$[x(1+r+r^2)]^2=[\frac{7}{3}]^2$

$x^2(1+2r+3r^2+2r^3+r^4)=\frac{49}{9}$ --- 3

so now 3/2

$
\frac{x^2(1+2r+3r^2+2r^3+r^4)}{x^2(1+r^2+r^4)}=\fr ac{49}{9}\times\frac{9}{91}
$

$21r^4+91r^3+112r^2+91r+21=0$

Since its a quartic ,

$

21(r^2+\frac{1}{r^2})+91(r+\frac{1}{r})+112=0
$

Let $r+\frac{1}{r}=p$

$21(p^2-2)+91p+112=0\Rightarrow$ p=-1(ommitted) , - $\frac{10}{3}$

therefore $p=-\frac{10}{3}$

$r+\frac{1}{r}=-\frac{10}{3}$

r=-3 , -1/3

so now head back to find x .

6. Oh, i substituted before squaring...^^