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

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

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$p^2, q^2, r^2$ are in $A.P.$

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(B) Given that $\frac{1}{p + q}, \frac{1}{r + p}, \frac{1}{q + r}$ are in $A.P.$Therefore,the common difference is equal:$\frac{1}{r + p} - \frac{1}{p + q} = \frac{1}{q + r} - \frac{1}{r + p}$Simplify the expressions:$\frac{(p + q) - (r + p)}{(r + p)(p + q)} = \frac{(r + p) - (q + r)}{(q + r)(r + p)}$$\frac{q - r}{p + q} = \frac{p - q}{q + r}$Cross-multiply:$(q - r)(q + r) = (p - q)(p + q)$$q^2 - r^2 = p^2 - q^2$$2q^2 = p^2 + r^2$This condition implies that $p^2, q^2, r^2$ are in $A.P.$

If $\frac{1}{p + q}, \frac{1}{r + p}, \frac{1}{q + r}$ are in $A.P.$,then

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