InterviewSolution
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InterviewSolution
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If `b-c,2b-lambda,b-a " are in HP, then " a-(lambda)/(2),b-(lambda)/(2),c-(lambda)/(2)` are isA. APB. GPC. HPD. None of these |
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Answer» Correct Answer - B `(2b-lambda)=(2(b-c)(b-a))/((b-c)+(b-a))` `implies (2b-lambda)=(2b-(a+c))=2[b^(2)-(a+c)b+ac]` `implies 2b^(2)-2blambda+lambda (a+c)-2ac=0` `implies b^(2)-blambda+(lambda)/(2)(a+c)-ac=0` `implies (b-(lambda)/(2))^(2)-lambda^(2)/(4)+lambda/(2)(a+c)-ac=0` `implies (b-(lambda)/(2))^(2)=lambda^(2)/(4)-(lambda)/(2)(a+c)+ac` `implies (b-(lambda)/(2))^(2)=(a-lambda/(2))(c-(lambda)/(2))` Hence, `a-(lambda)/(2),b-(lambda)/(2),c-(lambda)/(2)` are in GP. |
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