Given a + d b + c where a,;b,;c,;d are r | vedclass

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(b) $a + d > b + c$

$ \Rightarrow $ $a + b + c + d > 2b + 2c$

$ \Rightarrow $ $\frac{{a + c}}{2} + \frac{{b + d}}{2} > b + c$

$\frac

Vedclass · Accepted Answer

$\frac{1}{a},\;\frac{1}{b},\;\frac{1}{c},\;\frac{1}{d}$ are in $A.P.$

Vedclass · Answer

(b) $a + d > b + c$

$ \Rightarrow $ $a + b + c + d > 2b + 2c$

$ \Rightarrow $ $\frac{{a + c}}{2} + \frac{{b + d}}{2} > b + c$

$\frac{{a + c}}{2} > b$ and $\frac{{b + d}}{2} > c$ $[\because \;A > H]$

$b$ is the $H.M.$ of $a$ and $c$ and their $A.M.$ is $\frac{{a + c}}{2}$.

$c$ is $H.M.$ of $b$ and $d$ and their $A.M.$ is $\frac{{b + d}}{2}$.

Hence, $a,\;b,\;c,\;d$ are in $H.P.$

$ \Rightarrow $ $\frac{1}{a},\;\frac{1}{b},\;\frac{1}{c},\;\frac{1}{d}$ are in $A.P.$

Given $a + d > b + c$ where $a,\;b,\;c,\;d$ are real numbers, then

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