Browse Questions

If 4 positive numbers a, b, c and d are in HP, which of the following is true.

$\begin{array}{1 1} (a+c)^2 > 4ac \\ (b_d)^2 >4bd \\ both(a)\;and\;b \\ none\;of\;the\;above \end{array}$

Answer : (c) $both\;(a)\;\xi\;(b)$
Explanation: with any two numbers their AM >HM
Since a,b,c,d are in HM
$HM\; of\; a\;\xi\;c\qquad b=\frac{2ac}{a+c}$
$HM\;of\;b\;\xi\;d\qquad c=\frac{2bd}{b+d}$
$AM\; of\;a\;\xi\;c\;=\frac{a+c}{2}$
$AM\;of\;b\;\xi\;d\;=\frac{b+d}{2}$
$AM\;>\;HM$
$\frac{a+c}{2}\;>\;\frac{2ac}{a+c}\;\xi\;\frac{b+d}{2}\;>\;\frac{2bd}{b+d}$
$(a+c)^2\;>\;4ac\;\xi\;(b+d)^2\;>\;4bd$
$both\;of\;the\;inequalities\;are\;true.$