If `b-c,2b-lambda,b-a " are in HP, then " a-(lambda)/(2),b-(lambda)/(2),c-(lambda)/(2)` are is

Options

A. AP
B. GP
C. HP
D. None of these

Answer

B

Explanation

`(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.