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 :
[b]: Since\[\frac{1}{p+q},\frac{1}{r+p}\]and\[\frac{1}{r+q}\]are in A.P. \[\therefore \]\[\frac{1}{r+p}-\frac{1}{p+q}=\frac{1}{q+r}-\frac{1}{r+p}\] \[\Rightarrow \]\[\frac{p+q-r-p}{(r+p)(p+q)}=\frac{r+p-q-r}{(q+r)(r+p)}\] \[\Rightarrow \]\[\frac{q-r}{p+q}=\frac{p-q}{q+r}\]or\[{{q}^{2}}-{{r}^{2}}={{p}^{2}}-{{q}^{2}}\] Hence,\[{{p}^{2}},{{q}^{2}},{{r}^{2}}\]are in A.P.
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