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Q.

If x=n=0an,   y=n=0bn,   z=n=0cn, where a, b, c are in A.P. and |a|<1,|b|<1,|c|<1, abc 0, then 

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a

x, y, z are in A.P.
 

b

x, y, z are in G.P.
 

c

1x,1y,1z are in A.P.
 

d

1x+1y+1z=1-(a+b+c)

answer is C.

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Detailed Solution

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x=1+a+a2=. x=11-aa=1-1x y=11-bb=1-1y z=11-cc=1-1z a,b,careinA.P. 1-1x,1-1y,1-1zareinA.P. -1x,-1y,-1zareinA.P. 1x,1y,1zare inA.P.

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