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If a, b, c are in GP, prove that (a2 + b2 ), (ab + bc), (b2 + c2 ) are in GP. |
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Answer» To prove: (a2 + b2 ), (ab + bc), (b2 + c2 ) are in GP Given: a, b, c are in GP Formula used: When a,b,c are in GP, b2 = ac Proof: When a,b,c are in GP, b2 = ac … (i) Considering (a2 + b2 ), (ab + bc), (b2 + c2 ) (ab + bc)2 = (a2b2 + 2ab2c + b2c2 ) = (a2b2 + ab2c + ab2c + b2c2 ) = (a2b2 + b4 + a2c2 + b2c2 ) [From eqn. (i)] = [b2 (a2 + b2 )+ c2 (a2 + b2 )] (ab + bc)2 = [(b2 + c2 ) (a2 + b2 )] From the above equation we can say that (a2 + b2), (ab + bc), (b2 + c2) are in GP |
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