If frac{1}{b + c}, frac{1}{c + a}, frac{1}{a | vedclass

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(D) Given that $\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$ are in $AP$. This implies that $b + c, c + a, a + b$ are in Harmonic Progression $(

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None of these

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(D) Given that $\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$ are in $AP$. This implies that $b + c, c + a, a + b$ are in Harmonic Progression $(HP)$. Subtracting $a + b + c$ from each term,we get $(b + c) - (a + b + c), (c + a) - (a + b + c), (a + b) - (a + b + c)$ are in $HP$. This simplifies to $-a, -b, -c$ are in $HP$. Multiplying by $-1$,we get $a, b, c$ are in $HP$. If $a, b, c$ are in $HP$,then $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $AP$. However,the question asks for the progression of $a^2, b^2, c^2$. Since $a, b, c$ are in $HP$,there is no standard progression for $a^2, b^2, c^2$ unless specific values are given. Given the standard nature of this problem,if $a, b, c$ are in $HP$,then $a^2, b^2, c^2$ do not form a standard $AP, GP,$ or $HP$. Therefore,the correct option is $D$.

If $\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$ are in Arithmetic Progression $(AP)$,then $a^2, b^2, c^2$ are in which progression?

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