| (i) \[\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}\] are in A.P. |
| (ii) \[\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}\] are in A.P. |
A) (i) and (ii) both correct
B) (i) and (ii) both incorrect
C) (i) correct (ii) incorrect
D) (i) incorrect (ii) correct
Correct Answer: A
Solution :
| [a] Given \[{{a}^{2}},{{b}^{2}}.{{c}^{2}}\] are in A.P. |
| \[\Rightarrow {{a}^{2}}+(ab+bc+ca),{{b}^{2}}+(ab+bc+ca)\] |
| \[{{c}^{2}}+(ab+bc+ca)\] are in A.P. |
| \[\Rightarrow \,(a+b)(a+c),(b+c)(b+a),\,(c+a)(c+b)\] |
| are in A.P. |
| \[\Rightarrow \frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}\] are in A.P. |
| [Divide by \[(a+b)(b+c)(c+a)\]] |
| Again, \[{{a}^{2}},{{b}^{2}}.{{c}^{2}}\]are in A.P. |
| \[\Rightarrow \frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}\] are in A.P. |
| \[\Rightarrow \frac{a+b+c}{b+c},\frac{a+b+c}{c+a},\frac{a+b+c}{a+b}\] are in A.P. |
| \[\Rightarrow \frac{a}{b+c}+1,\frac{b}{c+a}+1,\frac{c}{a+b}+1\] are in A.P. |
| \[\Rightarrow \frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}\] are in A.P. |
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