If frac{1}{p+q}, frac{1}{r+p},and frac{1}{q+r | vedclass

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(B) Given that $\frac{1}{p+q}, \frac{1}{r+p}$,and $\frac{1}{q+r}$ are in $AP$.Therefore,the common difference is equal:$\frac{1}{r+p} - \frac{1}{p+q}

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$p^2, q^2, r^2$ are in $AP$.

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(B) Given that $\frac{1}{p+q}, \frac{1}{r+p}$,and $\frac{1}{q+r}$ are in $AP$.Therefore,the common difference is equal:$\frac{1}{r+p} - \frac{1}{p+q} = \frac{1}{q+r} - \frac{1}{r+p}$Simplifying the fractions:$\frac{(p+q) - (r+p)}{(r+p)(p+q)} = \frac{(r+p) - (q+r)}{(q+r)(r+p)}$$\frac{q-r}{p+q} = \frac{p-q}{q+r}$Cross-multiplying:$(q-r)(q+r) = (p-q)(p+q)$$q^2 - r^2 = p^2 - q^2$Rearranging the terms:$2q^2 = p^2 + r^2$This condition implies that $p^2, q^2, r^2$ are in $AP$.

If $\frac{1}{p+q}, \frac{1}{r+p}$,and $\frac{1}{q+r}$ are in Arithmetic Progression $(AP)$,then which of the following is true?

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