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If 1/a,1/b,1/c are in A.P and a,b -2c, a...

If `1/a,1/b,1/c` are in A.P and a,b -2c, are in G.P where a,b,c are non-zero then

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The correct Answer is:
A, B, D
`:. (1)/(a),(1)/(b),(1)/(c )` are in AP `implies a,b,c` are in HP
`:. B=(2ab)/(a+c)" ' "……(i)"`
and `a,b,-2c` are in GP, then `b^(2)=-2ac" " "….(ii)"`
From Eqs. (i) and (ii), we get
`b=(-b^(2))/(a+c) implies a+b+c=0 " " [:.b ne 0]`
`:. a^(3)+b^(3)+c^(3)=3abc` and `a,b,-2c` are in jGP
`implies a^(2),b^(2),4c^(2)` are also in GP and `a+b+c=0`
`implies 2b=-2a-2c`
`:.-2a,b,-2c` are in AP.
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