# If a,b,c are in AP and $a^2,b^2,c^2$ are in HP, then prove either $a=b=c$ or $a,b,- \frac c2$ are in GP

As the title says.

Although first part of the proof is obvious, I'm still able to prove it. And for the second part, I'm essentially trying to prove $b^2=-c/a$ (which is possible only when c<0 Xor a<0).

The relations found by me are: $a^2+c^2+2ac=4b^2$ and $b^2=\left(\frac {2a^2c^2}{a^2+c^2} \right)$.

Which provides me with:

$(a^2+c^2+2ac)/4=(2a^2c^2)/(a^2+c^2)$

I don't think this would lead me to the answer. Any help would be appreciated.

The AP condition gives $2b = a+c$ and the HP condition gives
$$\frac2{b^2}= \frac1{a^2}+\frac1{c^2} \iff \frac8{(a+c)^2}=\frac{a^2+c^2}{a^2c^2} \iff (a-c)^2(a^2+4ac+c^2)=0$$
Now either $(a-c)^2=0 \implies a=b=c$
or $a^2+4ac+c^2 = 0 \implies (a+c)^2+2ac=0 \implies 2b^2+ac=0 \implies a, b, -\dfrac{c}2$ are in GP.
• BTW, the last option is not feasible among real numbers, as $a,c$ having opposite signs gives $r^2<0$ where $r$ is the common ratio. Dec 17, 2014 at 7:29