Find $x+y+z$ given that $x, y , z$ are in an arithmetic series with ratio r, while $x, y+2, z+12$ are in a geometric series with ratio $r+1$.

Question

I am told that the positive numbers $x, y, z$ are in an arithmetic series with the ratio $r$ and that the numbers $x, y+2, z + 12$ are in a geometric series with ratio $r + 1$. I have to find the sum $x + y + z$, and the following options are given:

A. $12$

B. $-12$

C. $9$

D. $7$

E. $15$

What I tried is to use the properties of arithmetic and geometric series, so

• Given that $x, y, z$ are in an arithmetic series with the ratio $r$, we have:

$\hspace{1.5cm} y = x + r$

$\hspace{1.5cm} z = x + 2r$

• Given that $x, y+2, z+12$ are in an arithmetic series with the ratio $r+1$, we have:

$\hspace{1.5cm} y + 2 = x(r + 1) \Rightarrow y=xr+x-2$

$\hspace{1.5cm} z + 12 = x(r+1)^2 \Rightarrow z = xr^2+2xr+x-12$

I don't see how I can use these to find the sum $x+ y + z$, especially since I have to find the sum as a number and not the sum based on $x$'s and $r$'s.

Answer 1

We have $$x,y=x+r,z=x+2r$$ and $$x,y+2=x(r+1),z+12=x(r+1)^2$$ so we get $$x+r+2=x(r+1)$$ and $$x+2r+12=x(r+1)^2$$ eliminating $$r(x-1)=2$$ we get the equation $$x+\frac{4}{x-1}+12=x\left(\frac{2}{x-1}+1\right)^2$$ solving this we get $$x=2$$ or $$x=\frac{1}{2}$$

Answer 2

You can solve the four equations in four unknowns ($x,y,z, $ and $r$), as follows.

$z+12=x+2r+12=x(r+1)^2=xr^2+2xr+x,$ so $2r+12=xr^2+2xr$.

and $y+2=x+r+2=x(r+1)=xr+x,$ so $r+2=xr,$ so $2r+4=2xr$ .

Subtract to get $8=xr^2$.

Now substitute $x=\dfrac8{r^2}$ in $r+2=xr$ to get $r+2=\dfrac8r$ or $r^2+2r-8=(r-2)(r+4)=0$,

so $r=2$ or $r=-4$. But $r=-4$ is inconsistent with $x, y,$ and $z$ all being positive, so $r=2$.

Now you should be able to solve for $x, y, $ and $z$ and add them up.

Answer 3

$x+y+z=3a+3r$, so the answer is a multiple of $3$ and positive. $a+r\in\{3,4,5\}$.

$\dfrac{z+12}{y+2}=\dfrac{y+2}{x}$

So $(a+2r+12)a=(a+r+2)^2$

and $a^2+2ra+12a=a^2+r^2+4+2ar+4a+4r$

$8a=r^2+4r+4$

So $r=2k$:

$8a=4k^2+8k+4\implies 2a=k^2+2k+1\implies 2a=(k+1)^2\implies k=1, a=2, r=2$

Answer:A (12)

Answer 4

Hint

$x=y-r,z=y+r$

As $r+1\ne1$

$\dfrac{y+2}{y-r}=\dfrac{y+r+12}{y+2}=\dfrac{y+r+12-(y+2)}{y+2-(y-r)}=?$

which needs to be $r+1$