If a, b, c, d are in H.P.,then | vedclass

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(C) Given that $a, b, c, d$ are in $H.P.$Since $a, b, c$ are in $H.P.$,$b$ is the $H.M.$ between $a$ and $c$. We know that for any two positive number

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Both $(a)$ and $(b)$

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(C) Given that $a, b, c, d$ are in $H.P.$Since $a, b, c$ are in $H.P.$,$b$ is the $H.M.$ between $a$ and $c$. We know that for any two positive numbers,$A.M. > G.M. > H.M.$For $a$ and $c$,$A.M. = \frac{a+c}{2}$,$G.M. = \sqrt{ac}$,and $H.M. = b$.Since $A.M. > H.M.$,we have $\frac{a+c}{2} > b \Rightarrow a + c > 2b$ .... $(i)$Since $G.M. > H.M.$,we have $\sqrt{ac} > b \Rightarrow ac > b^2$ .... $(ii)$Similarly,for $b, c, d$ in $H.P.$,$c$ is the $H.M.$ between $b$ and $d$.Thus,$A.M. > H.M. \Rightarrow \frac{b+d}{2} > c \Rightarrow b + d > 2c$ .... $(iii)$And $G.M. > H.M. \Rightarrow \sqrt{bd} > c \Rightarrow bd > c^2$ .... $(iv)$Adding $(i)$ and $(iii)$,we get $(a + c) + (b + d) > 2b + 2c \Rightarrow a + d > b + c$.Multiplying $(ii)$ and $(iv)$,we get $(ac)(bd) > (b^2)(c^2) \Rightarrow abcd > b^2c^2 \Rightarrow ad > bc$.Therefore,both $(a)$ and $(b)$ are correct.

If $a, b, c, d$ are in $H.P.$,then

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