If positive numbers a, b, c, d are in harmonic progression and $a \neq b$ , then-

a + d > b + c is always true.

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a + b > c + d is always true.

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a + c > b + d always true.

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ad > bc

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Solution

The correct options are
C ad > bc
D a + d > b + c is always true.
Take $a = \frac{1}{p - 3 q}, b = \frac{1}{p - q}, c = \frac{1}{p + q}, d = \frac{1}{p + 3 q}$

Then $a + d > b + c$ easily follows

Since $(a + d) - (b + c)$

$= \frac{2 p}{p^{2} - 9 q^{2}} - \frac{2 p}{p^{2} - q^{2}} = 2 p [\frac{8 q^{2}}{(p^{2} - 9 q^{2}) (p^{2} - q^{2})}]$

Which is positive $(∵ a, b, c, d > 0)$

Also $a d - b c = \frac{1}{p^{2} - 9 q^{2}} - \frac{1}{p^{2} - q^{2}}$

$= \frac{8 q^{2}}{(p^{2} - 9 q^{2}) (p^{2} - q^{2})} > 0$ .
Hence, options 'A' and 'D' are correct.

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If positive numbers a, b, c, d are in harmonic progression and a≠b, then-

If positive numbers a, b, c, d are in harmonic progression and $a \neq b$ , then-