I found this problem recently:
Given that: $a, b, c$ form an arithmetic progression, $b, c, d$ form a geometric progression and $c,d,e$ form a harmonic progression, prove that $a,c,e$ form a geometric progression.
My thoughts: We have $a=a, b=a+k, c=a+2k$ (for some constant $k$), from which we get $$(1)\ 2b=a+c \ .$$
Also, we have that $b=b, c=mb, d=m^2b$, from which we get that $$(2) \ c^2=bd \ .$$
Finally, since $c,d,e$ are in HP, we have that $c=c, d=\frac1{c+n}, e=\frac1{c+2n}$ for some constant $n$.
Here are where my useful attempts end: just lots of algebraic manipulation that leads nowhere.
The solution starts off similar to mine, but then says that:
$$ (3) \ d=\frac{2ce}{c+e}$$ (I don't know where one would naturally deduce this), from which we get that $$2 \frac{c^2}{\frac{2ce}{c+e}}=a+c$$ which apparently implies that $$c^2=ae \ ,$$ which, apparently, implies that $a,c,e$ are in GP.
Could someone explain these last few steps to me? Am I missing something obvious?
Also, why does $c^2=ae$ imply that $a,c,e$ follow a geometric progression?