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Question

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

A
GP
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B
AP
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C
Arithmetic- Geometric progression
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D
H.P
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Solution

The correct option is D H.P
x=n=0an=11a or a=11xy=n=0bn=11b or b=11yz=n=0cn=11c or c=11z
a,b,c are in A.P.
2b=a+c
2(11y)=11x+11y2y=1x+1zx,y,z are in H.P.

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