Solve three variables that are consecutive terms in a geometric suite?

Question

Can somebody give me a hint on how to start solving this problem. I have no idea.

$x$, $y$ and $z$ are consecutive terms in a geometric suit satisfying $$x+y+z=\frac73\ \text{and}\ x^2+y^2+z^2=\frac{91}9$$

Answer 1

HINT

You know that $x$ $y$ and $z$ are CONSECUTIVE terms of a geometric suit, let say of number $q$. Can you express $y$ and $z$ in terms of $x$ and $q$ ?

Answer 2

$z = 7/3 - x - y$. Plugging it back in: $x^2 + y^2 + (7/3 - x - y)^2 = 91/9$ which is the equation of a circle. Then assume that there exists $(a,b)$ so that $x = a$ and $y = ab$ and $z=ab^2$ (definition of consecutive terms of a geometric sequence). Plug it back in, and you should reach the answer.

Answer 3

If $q$ is the ratio of the geometric sequence, and if $x,y,z$ are consecutive terms, we have $$ x=\frac yq,\quad z=yq.$$ The given relation can be written as $$x+y+z=\Bigl(q+\frac1q+1\Bigr)y=\frac 73,\quad x^2+y^2+z^2=\Bigl(q^2+\frac1{q^2}+1\Bigr)y^2=\frac{91}9.$$ Set $t=q+\dfrac1q$. The second equation becomes $$(t^2-1)y^2=\frac{91}9$$ and squaring the first we obtain $$(t+1)^2y^2=\frac{49}9,\enspace\text{whence} \quad\frac{t^2-1}{(t+1)^2}=\frac{t-1}{t+1}=\frac{13}7.$$ (we supposed $t\ne -1$, which would correspond to a complex $q$).

Can you end the computations?

Answer 4

$$x+y+z=\frac73\ \text{and}\ x^2+y^2+z^2=\frac{91}9\tag{1}$$ Multiply through by $3$ and $9$ respectively in $(1)$ to give $$3x+3y+3z=7\ \text{and}\ (3x)^2+(3y)^2+(3z)^2=91\tag{2}$$

Square the first equation in $(2)$ \begin{align*} (3 x + 3 y + 3 z)^2&=(3x)^2+(3y)^2+(3z)^2 + 18 x y + 18 x z + 18 y z \\ &=91+ 18 x y + 18 x z + 18 y z \quad\text{($91$ from the $2$nd eqn. in $(2)$)} \\ &=7^2=49\\ \end{align*} Therefore $$18 x y + 18 x z + 18 y z =49-91=-42\implies x y + x z + y z=-\frac73$$

Now if $x$, $y$ and $z$ are in a GS then $\frac{x}{y}=\frac{y}{z}$, so $y^2=xz$ So $$ x y + x z + y z= x y + y^2 + y z=y(x+y+z)=-\frac73$$ But we know $x+y+z=\frac73$, so $y=-1$. Therefore $y^2=xz=1$

Now $$x+z=1+\frac73=\frac{10}{3}\ \text{and}\ x z=1\implies z=\frac1x\implies x+\frac1x=\frac{10}{3}$$ Therefore $$3x^2-10x+3=0\implies (3x-1)(x-3)=0\implies \ x=\frac13\ \text{or}\ x=3$$ Now $z=\frac1x$, so $z=3$ or $z=\frac13$ respectively. So by symmetry one of $x$ or $z$ is $3$ and the other $\frac13$.

Answer 5

$$z=\frac73-x-y\ \text{so}\ z^2=\left(\frac73-x-y\right)^2=\frac{49}{9}-\frac{14}{3}x+x^2-\frac{14}{3}y+2xy+y^2\tag{1}$$ Substituting this into $x^2+y^2+z^2$ and simplifying we get the equation of an ellipse

\begin{align*} \left(x-\frac76\right)^2+\left(y-\frac76\right)^2+xy-\frac{91}{18}=0 \end{align*}

which simplifies to $$3x^2 + (3y - 7)x + (3y^2 - 7y - 7)=0$$ a quadratic in $x$ with solutions $$x=\frac{(7-3y)\pm\sqrt{-27 y^2 + 42 y + 133}}{6}$$

We need $-27 y^2 + 42 y + 133=d^2$, for some $d\in\Bbb{Z}$. Complete the square $$ -27 \left[y^2 - \frac{42}{27} y - \frac{133}{27}\right] =-27 \left[\left(y - \frac{7}{9}\right)^2 - \frac{448}{81}\right]=d^2 $$ Rearrange \begin{align*} \left[\left(y - \frac{7}{9}\right)^2 - \frac{448}{81}\right]=-\frac{d^2}{27}\implies y=\frac{7\pm\sqrt{448-3d^2} }{9} \end{align*} At $d=4$ we find $448-3\cdot4^2=400$, with $\sqrt{400}=20$. This gives $y=\frac{7+20 }{9}=3$ or $y=\frac{7-20 }{9}=\frac{13}{9}$ Now try the different values to find $x$ and $z$ in geometric progression. Try $y=3$, then $x_+=\frac13$, $x_-=-1$. Pick $x_+$. Now we know $x+y+z=\frac73$, so try $z=\frac73-3-\frac13=-1$ and we find $x=\frac13$, $z=-1$, and $y=3$ in GP.