A) A.P.
B) GP.
C) H.P.
D) None of these
Correct Answer: A
Solution :
| \[\because \,\,{{a}^{2}},{{b}^{2}},{{c}^{2}}\]are in A.P. |
| \[\therefore \,\,{{a}^{2}}+ab+bc+ca,\]\[{{b}^{2}}+bc+ca+ab,\] \[{{c}^{2}}+ca+ab+bc\] ..... are also in A.P. [adding \[ab+bc+ca\]] |
| or \[\left( a+c \right)\,\left( a+b \right),\] \[\left( b+c \right)\left( a+b \right),\] \[\left( c+a \right)\left( b+c \right)\].. are also |
| in A.P. \[\Rightarrow \,\,\,\frac{1}{b+c},\,\frac{1}{c+a},\,\frac{1}{a+b}\] are in A.P. |
| [dividing by \[(a+b)\,(b+c)\,(c+a)\]] |
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