Slow and Steady wins the race...

To start, I will use a property of triplets in arithmetic progression: $a\Big(\frac{1}{b}+\frac{1}{c}\Big)+c\Big(\frac{1}{a}+\frac{1}{b}\Big)=2b\Big(\frac{1}{c}+\frac{1}{a}\Big).$ Multiply through by $abc$ , and rearrange the terms to get $2(a+c)b^2-(a^2+c^2)b-ac(a+c)=0\Rightarrow b=\frac{a^2+c^2\pm\sqrt{a^4+2a^2c^2+b^4-8(-ac)(a+c)^2}}{4(a+c)}$ $\Rightarrow b=\frac{a^2+c^2\pm\sqrt{a^4+8a^3c+16a^2c^2+8ac^3+c^4}}{4(a+c)}=\frac{a^2+c^2\pm (a^2+4ac+c^2)}{4(a+c)}=\frac{a+c}{2},-\frac{ac}{a+c}$ The solution $b=-\frac{ac}{a+c}$ implies $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ , which, in turn, implies the common difference is $b\Big(\frac{1}{c}+\frac{1}{a}\Big)-a\Big(\frac{1}{b}+\frac{1}{c}\Big)=b\Big(-\frac{1}{b}\Big)-a\Big(-\frac{1}{a}\Big)=-1+1=0$ . So, for example, $a=1,b=-2,c=-2$ gives a constant sequence for $a\Big(\frac{1}{b}+\frac{1}{c}\Big),b\Big(\frac{1}{c}+\frac{1}{a}\Big),c\Big(\frac{1}{a}+\frac{1}{b}\Big)$ because $1-\frac{1}{2}-\frac{1}{2}=0$ , but the sequence $a,b,c$ is clearly not arithmetic. So, I must add the assumption that the original sequence is not constant. The other solution $b=\frac{a+c}{2}$ is equivalent to $a+c=2b$ , which allows us to conclude $a,b,c$ are in arithmetic progression.