Correct Answer - B
It is given that
b-c, 2b-x, b-a are in H.P.
`rArr" "(1)/(b-c),(1)/(2b-x),(1)/(b-a)` are in A.P.
`rArr" "(2)/(2b-x)=(1)/(b-c)+(1)/(b-a)`
`rArr" "(1)/(b-(x)/(2))=(1)/((b-(x)/(2))-(c-(x)/(2)))+(1)/((b-(x)/(2))-(a-(x)/(2)))`
`rArr" "(1)/(B)=(1)/(B-C)+(1)/(B-A)," where "A=a-(x)/(2),B=b-(x)/(2),C=c-(x)/(2)`
`rArr" "B^(2)-B(C+A)+CA-2B^(2)-B(A+C)`
`rArr" "B^(2)=ACrArrA,B,C` are in G.P.
`rArr" "a-(x)/(2),b-(x)/(2)andc-(x)/(2)` are in G.P.
