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Step-by-step Solution
Step 1: Recall the relationships for roots of a quadratic
Consider the quadratic equation $a x^2 + b x + c = 0$, with roots $ \alpha $ and $ \beta $. By Viète's relations:
Sum of roots: $ \alpha + \beta = -\frac{b}{a}$
Product of roots: $ \alpha \beta = \frac{c}{a}$
Step 2: Express the given condition in mathematical form
The problem states that the sum of the roots equals the sum of the squares of their reciprocals:
$$
\alpha + \beta
\;=\;
\frac{1}{\alpha^2} \;+\; \frac{1}{\beta^2}.
$$
Step 3: Rewrite the sum of the squares of the reciprocals
The sum of squares of the reciprocals may be written as
$$
\frac{1}{\alpha^2} + \frac{1}{\beta^2}
\;=\;
\frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2}.
$$
Next, recall that
$$
\alpha^2 + \beta^2
\;=\;
(\alpha + \beta)^2 - 2 \alpha \beta.
$$
Step 4: Substitute known expressions into the condition
Using $ \alpha + \beta = -\frac{b}{a}$ and $ \alpha \beta = \frac{c}{a}$, we get
$$
\alpha^2 + \beta^2
\;=\;
\left( -\frac{b}{a} \right)^2 - 2 \left( \frac{c}{a} \right)
\;=\;
\frac{b^2}{a^2} - \frac{2c}{a}.
$$
Hence,
$$
\frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2}
\;=\;
\frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\left(\frac{c}{a}\right)^2}
\;=\;
\frac{\frac{b^2 - 2ac}{a^2}}{\frac{c^2}{a^2}}
\;=\;
\frac{b^2 - 2ac}{c^2}.
$$
Therefore the given condition becomes:
$$
\alpha + \beta
\;=\;
\frac{b^2 - 2ac}{c^2}.
$$
But we already have $ \alpha + \beta = -\frac{b}{a} $. Equating these:
$$
-\frac{b}{a}
\;=\;
\frac{b^2 - 2ac}{c^2}.
$$
Step 5: Simplify to find the relationship among $a$, $b$, and $c$
Cross-multiplying gives:
$$
-b \, c^2 = a \left( b^2 - 2ac \right).
$$
Rearranging leads to:
$$
a b^2 + b c^2 = 2 a^2 c.
$$
Divide throughout by $b$ (assuming $b \neq 0$):
$$
\frac{a b^2}{b} + \frac{b c^2}{b}
\;=\;
\frac{2 a^2 c}{b}
\quad\Longrightarrow\quad
a b + c^2
\;=\;
2 \frac{a^2 c}{b}.
$$
Or equivalently:
$$
2 \frac{a^2 c}{b}
\;=\;
a b + c^2
\quad\Longrightarrow\quad
2 \frac{a}{b}
\;=\;
\frac{c}{a} + \frac{b}{c}.
$$
Step 6: Recognize the pattern of the progressions
From
$$
2 \frac{a}{b}
\;=\;
\frac{c}{a} + \frac{b}{c},
$$
it follows that $ \frac{c}{a}, \; \frac{a}{b}, \; \frac{b}{c} $ are in arithmetic progression (A.P.). Consequently, taking reciprocals shows that
$$
\frac{a}{c}, \; \frac{b}{a}, \; \frac{c}{b}
$$
lie in harmonic progression (H.P.).
Conclusion
Therefore, $ \frac{a}{c}, \; \frac{b}{a}, \; \frac{c}{b} $ are in harmonic progression, as was to be shown.
