A) \[p,q,r\]will be in AP
B) \[\frac{1}{p},\frac{1}{q},\frac{1}{r}\]will be in AP
C) \[{{p}^{2}},{{q}^{2}},{{r}^{2}}\]will be in AP
D) \[\frac{1}{{{p}^{2}}},\frac{1}{{{q}^{2}}},\frac{1}{{{r}^{2}}}\]will be in AP
Correct Answer: C
Solution :
If\[\frac{1}{q+r},\frac{1}{r+p},\frac{1}{p+q}\]are in AP. Then, \[\frac{1}{r+p}-\frac{1}{q+r}=\frac{1}{p+q}-\frac{1}{r+p}\] \[\Rightarrow \]\[\frac{q+r-r-p}{(r+p)(q+r)}=\frac{r+p-p-q}{(p+q)(r-p)}\] \[\Rightarrow \]\[\frac{(p-q)}{(q+r)}=\frac{(r-q)}{(p+q)}\] \[\Rightarrow \]\[(q-p)(q+p)=(r-q)(r+q)\] \[\Rightarrow \]\[{{q}^{2}}-{{p}^{2}}={{r}^{2}}-{{q}^{2}}\] \[\Rightarrow \]\[{{p}^{2}},{{q}^{2}},{{r}^{2}}\]are in AP.
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