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If x ,1 ,z are in A.P. and x , 2 , z are in G.P. then x , 4 , z are in ?

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Science master

1)2=x+z 2)4=xz Divide 1 by 2 2/4=x+z/xz 2(1/4)=1/x+1/z =>1/4,1/x,1/z in A.P so,4,x,z in H.P
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Maths(concepts clarity is my priority)

Hp sir
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AP
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Math Magician

They are in H.P. because 1/x,1/4 and 1/z are in A P. As 1/x+1/z =(x+z)/xz =2/4 (x+z =2 from A P and xz=4 from GP expression given.) Which is equal to 2/4 . so, x, 4, z are in HP
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Math Magician

They are in H.P. because 1/x,1/4 and 1/z are in A P. As 1/x+1/z =(x+z)/xz =2/4 (x+z =2 from A P and xz=4 from GP expression given.) Which is equal to 2/4 . so, x, 4, z are in HP.
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Math Magician

HP because 1/x, 1/4, 1/z are in AP.As x+z =2 from AP expression given and xz =4 from GP condition given as 1/x+1/z= (x+z) / xz = 2/4 (from analyzing the given conditions) so, 1/x+1/z=2/4 i.e. 2/2nd term. thus, the terms are in HP
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HP
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x, 4 , z are in H.P. ( harmonic progression)
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