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

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(C) If $a, b, c, d$ are in $H.P.$,then their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in $A.P.$For any four terms in $H.P.

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

Vedclass · Answer

(C) If $a, b, c, d$ are in $H.P.$,then their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in $A.P.$For any four terms in $H.P.$,we know that $a, b, c$ are in $H.P.$,so $b = \frac{2ac}{a+c}$. Since $A.M. > H.M.$,we have $\frac{a+c}{2} > b$,which implies $a+c > 2b$.Similarly,for $b, c, d$ in $H.P.$,we have $b+d > 2c$.Adding these two inequalities: $(a+c) + (b+d) > 2b + 2c$,which simplifies to $a+d > b+c$. Thus,$(a)$ is true.Also,for $a, b, c$ in $H.P.$,$ac > b^2$. For $b, c, d$ in $H.P.$,$bd > c^2$.Multiplying these: $(ac)(bd) > (b^2)(c^2)$,which simplifies to $ad > bc$. Thus,$(b)$ is true.Therefore,both $(a)$ and $(b)$ are correct.

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

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