If a, b, c, d are four numbers such that the first three are in AP while the last three are in HP then

bc=ad

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ac=bd

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ab=cd

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none of these

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Solution

The correct option is A bc=ad
Given
$2 b = a + c$ and
$\frac{2}{c} = \frac{1}{b} + \frac{1}{d}$
$\Rightarrow \frac{2}{c} - \frac{2}{a + c} = \frac{1}{d}$
$\Rightarrow 2 (\frac{a + c - c}{c (a + c)}) = \frac{1}{d}$
$\Rightarrow \frac{a}{c b} = \frac{1}{d}$
$∴ a d = b c$

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If a, b, c, d are four numbers such that the first three are in AP while the last three are in HP then

The correct option is A bc=adGiven 2b=a+c and2c=1b+1d⇒2c−2a+c=1d⇒2(a+c−cc(a+c))=1d⇒acb=1d∴ad=bc

The correct option is A bc=ad
Given
$2 b = a + c$ and
$\frac{2}{c} = \frac{1}{b} + \frac{1}{d}$
$\Rightarrow \frac{2}{c} - \frac{2}{a + c} = \frac{1}{d}$
$\Rightarrow 2 (\frac{a + c - c}{c (a + c)}) = \frac{1}{d}$
$\Rightarrow \frac{a}{c b} = \frac{1}{d}$
$∴ a d = b c$