If x, 1, z are in A.P. and x, 2, z are in G.P | vedclass

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(C) Given that $x, 1, z$ are in $A.P.$,the middle term is the arithmetic mean: $1 = \frac{x + z}{2}$,which implies $x + z = 2$......$(i)$Given that $x

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$H.P.$

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(C) Given that $x, 1, z$ are in $A.P.$,the middle term is the arithmetic mean: $1 = \frac{x + z}{2}$,which implies $x + z = 2$......$(i)$Given that $x, 2, z$ are in $G.P.$,the middle term is the geometric mean: $2^2 = xz$,which implies $xz = 4$......$(ii)$For $x, 4, z$ to be in $H.P.$,the middle term must be the harmonic mean of $x$ and $z$,which is $\frac{2xz}{x + z}$.Substituting the values from $(i)$ and $(ii)$ into the expression for the harmonic mean:$	ext{Harmonic Mean} = \frac{2(4)}{2} = \frac{8}{2} = 4$.Since the middle term is $4$,$x, 4, z$ are in $H.P.$

If $x, 1, z$ are in $A.P.$ and $x, 2, z$ are in $G.P.$,then $x, 4, z$ will be in

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