Arithmetic and Geometric Sequences Together

Since $x$ , $y$ , and $z$ are in a geometric sequence, we can express $y$ as $xr$ and $z$ as $xr^2$ . Therefore, since $3x$ , $26y$ , and $17z$ are in an arithmetic progression, we can construct the equation $17z-26y=26y-3x$ . We can plug in the values of $y$ and $z$ in the equation to get $17xr^2-26xr=26xr-3x.$ We can eliminate the $x$ terms to get $17r^2-26r=26r-3.$ From this, we can rearrange terms and get the quadratic $17r^2-52r+3=0.$ By Vieta's, the sum of the roots of a quadratic is equivalent to $-\dfrac{b}{a}$ , so the sum of all possible values of $r$ is $\dfrac{52}{17}$ . Our answer is $52+2(17)=\boxed{86}$ .

I did the same.