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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.

Harshal Gajjar
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1 Answers1

4

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.

Macavity
  • 46,381
  • 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. – Macavity Dec 17 '14 at 07:29