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 $\displaystyle \sum_{k=1}^n ar^k$
So $\displaystyle x=x$, and y in the next term so $\displaystyle y=rx$ and $\displaystyle z=ry=r^2x$
So $\displaystyle x+rx+r^2x=\frac{7}{3}=x(1+r+r^2)$
And the second equation tells us
$\displaystyle
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
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$\displaystyle x(1+r+r^2)=\frac{7}{3}$ --- 1
$\displaystyle x^2(1+r^2+r^4)=\frac{91}{9} $ --- 2
From 1
$\displaystyle [x(1+r+r^2)]^2=[\frac{7}{3}]^2$
$\displaystyle x^2(1+2r+3r^2+2r^3+r^4)=\frac{49}{9}$ --- 3
so now 3/2
$\displaystyle
\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}
$
$\displaystyle 21r^4+91r^3+112r^2+91r+21=0$
Since its a quartic ,
$\displaystyle
21(r^2+\frac{1}{r^2})+91(r+\frac{1}{r})+112=0
$
Let $\displaystyle r+\frac{1}{r}=p$
$\displaystyle 21(p^2-2)+91p+112=0\Rightarrow$ p=-1(ommitted) , -$\displaystyle \frac{10}{3}$
therefore $\displaystyle p=-\frac{10}{3}$
$\displaystyle r+\frac{1}{r}=-\frac{10}{3}$
r=-3 , -1/3
so now head back to find x .