The sum of the first three members of a geometric progression is equal to 42. Those same members are, correspondingly, the first, the second and the sixth member of an arithmetic progression. The task is to find out the values of those three elements. (Solution: 2, 8, 32 or 14, 14, 14). After a few lines of solving, I got an equation that describes the difference in the arithmetic progression as a function of x: $$f(x)=(14-2x)(1+\frac{14-x}{14-2x}+\frac{(14-x)^2}{(14-2x)^2})$$ $$y=42$$ I'm 100% sure I overcomplicated this, but now I'd really like to know how to get the intersection points of this weird function with a line (the sum). After plotting the equation in Desmos, I get two solutions; 0 and 6. Those are indeed the appropriate differences from which I can obtain the three numbers, but I have no idea how to get to them.
Any help/question reformatting is appreciated!
Let your numbers be $x,rx,r^2x$. Since $x+rx+r^2x=42$, $x\neq 0$. Also,
$$r^2x-x=(r^2-1)x=5(rx-x)=5x(r-1) \implies r^2-1=5(r-1) \implies (r-1)(r-4)=0.$$ Therefore, $r=1$ or $r=4$. Now use $x+rx+r^2x=42$ to find the sequence.