A) \[p,\ ,q,\ r\] are in A.P.
B) \[{{p}^{2}},\ {{q}^{2}},\ {{r}^{2}}\] are in A.P.
C) \[\frac{1}{p},\ \frac{1}{q},\ \frac{1}{r}\] are in A.P.
D) None of these
Correct Answer: B
Solution :
Since \[\frac{1}{p+q},\ \frac{1}{r+q}\] and \[\frac{1}{q+r}\] are in A.P. \[\therefore \]\[\frac{1}{r+q}-\frac{1}{p+q}=\frac{1}{q+r}-\frac{1}{r+p}\] \[\Rightarrow \] \[\frac{p+q-r-p}{(r+p)(p+q)}=\frac{r+p-q-r}{(q+r)(r+p)}\] \[\Rightarrow \] \[\frac{q-r}{p+q}=\frac{p-q}{q+r}\] or \[{{q}^{2}}-{{r}^{2}}={{p}^{2}}-{{q}^{2}}\] \[\therefore \] \[2{{q}^{2}}={{r}^{2}}+{{p}^{2}}\] Therefore \[{{p}^{2}},\ {{q}^{2}},\ {{r}^{2}}\] are in A.P.
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