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If `(a^(2)+b^(2)), (ab+bc), (b^(2)+c^(2))` are in GP then prove that a, b, c are also in GP. |
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Answer» Let `(a^(2)+b^(2)), (ab+bc)` and `(b^(2)+c^(2))` be in GP. Then, `(ab+bc)^(2)=(a^(2)+b^(2))(b^(2)+c^(2))` `rArr a^(2)b^(2)+b^(2)c^(2)+2ab^(2)c=a^(2)b^(2)+a^(2)c^(2)+b^(4)+b^(2)c^(2)` `rArr b^(4)+a^(2)c^(2)-2ab^(2)C=0` `rArr (b^(2)-ac)^(2)=0 rArr b^(2)-ac=0 rArr b^(2)=ac`. Hence, a, b, c are in GP. |
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