1. Let the roots of $ax^2+bx+c=0$ be $\alpha$ and $\beta$.

Then by Vieta's formulas, $\alpha+\beta=-\frac{b}{a}, \qquad \alpha\beta=\frac{c}{a}.$

2. Given condition: $\alpha+\beta=\frac{1}{\alpha^2}+\frac{1}{\beta^2}.$

Now, $\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}.$ Also, $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta.$

So the condition becomes $\alpha+\beta=\frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}.$

3. Put $S=\alpha+\beta=-\frac{b}{a}, \qquad P=\alpha\beta=\frac{c}{a}.$ Then $S=\frac{S^2-2P}{P^2}.$ Thus, $SP^2=S^2-2P.$

4. Substitute $S=-\frac{b}{a}$ and $P=\frac{c}{a}$: $\left(-\frac{b}{a}\right)\left(\frac{c}{a}\right)^2=\left(-\frac{b}{a}\right)^2-2\left(\frac{c}{a}\right).$

Simplify: $-\frac{bc^2}{a^3}=\frac{b^2}{a^2}-\frac{2c}{a}.$ Multiply by $a^3$: $-bc^2=ab^2-2a^2c.$ Rearrange: $ab^2+bc^2=2a^2c.$ Factor $b$ on the left: $b(ab+c^2)=2a^2c.$ But a more useful form is $b(a^2+c^2)=2abc$ only if we derive carefully by another direct substitution. Let us verify properly.

5. Instead, substitute directly into $S=\frac{S^2-2P}{P^2}.$

We have $S^2-2P=\frac{b^2}{a^2}-\frac{2c}{a}=\frac{b^2-2ac}{a^2},$ and $P^2=\frac{c^2}{a^2}.$ Hence $\frac{S^2-2P}{P^2}=\frac{b^2-2ac}{c^2}.$ So $-\frac{b}{a}=\frac{b^2-2ac}{c^2}.$ Multiply by $ac^2$: $-bc^2=a(b^2-2ac)=ab^2-2a^2c.$ Thus, $ab^2+bc^2=2a^2c.$ Factor $b$: $b(ab+c^2)=2a^2c.$

6. Now check the sequence $\frac{a}{c},\quad \frac{b}{a},\quad \frac{c}{b}.$ To be in harmonic progression, their reciprocals must be in arithmetic progression.

Their reciprocals are $\frac{c}{a},\quad \frac{a}{b},\quad \frac{b}{c}.$ For these to be in A.P., we need $2\cdot \frac{a}{b}=\frac{c}{a}+\frac{b}{c}.$ Multiply by $abc$: $2a^2c=bc^2+ab^2.$ This is exactly the relation obtained above: $ab^2+bc^2=2a^2c.$

Hence, $\frac{a}{c},\ \frac{b}{a},\ \frac{c}{b}$ are in Harmonic Progression.

7. Therefore the correct option is $\boxed{\text{D}}.$