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

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(b) Since $\frac{1}{{p + q}},\;\frac{1}{{r + q}}$ and $\frac{1}{{q + r}}$ are in $A.P.$

$	herefore $ $\frac{1}{{r + q}} - \frac{1}{{p + q}} = \fra

Vedclass · Accepted Answer

${p^2},\;{q^2},\;{r^2}$ are in $A.P.$

Vedclass · Answer

(b) Since $\frac{1}{{p + q}},\;\frac{1}{{r + q}}$ and $\frac{1}{{q + r}}$ are in $A.P.$

$	herefore $ $\frac{1}{{r + q}} - \frac{1}{{p + q}} = \frac{1}{{q + r}} - \frac{1}{{r + p}}$

$ \Rightarrow $ $\frac{{p + q - r - p}}{{(r + p)(p + q)}} = \frac{{r + p - q - r}}{{(q + r)(r + p)}}$

$ \Rightarrow $ $\frac{{q - r}}{{p + q}} = \frac{{p - q}}{{q + r}}$ or ${q^2} - {r^2} = {p^2} - {q^2}$

$	herefore $ $2{q^2} = {r^2} + {p^2}$

Therefore ${p^2},\;{q^2},\;{r^2}$ 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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