If a- b- c are in AP- then 1-bc- 1-ca and 1-ab are in AP-

If a, b, c are in AP, then a +c = 2b .... (i) To check if 1/bc, 1/ca and 1/ab are in AP, 1/bc + 1/ab = a + c/abc = 2b/abc = 2/ac Hence proved. So the sta ...

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Standard X

Mathematics

Arithmetic Progression

If a, b, c ar...

Question

If $a, b, c$ are in AP; then $\frac{1}{b c}, \frac{1}{c a} a n d \frac{1}{a b}$ are in AP.

True

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False

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Solution

The correct option is A
True

If $a, b, c$ are in AP, then a +c = 2b .... (i)

To check if $\frac{1}{b c}, \frac{1}{c a} a n d \frac{1}{a b}$ are in AP,

$\frac{1}{b c} + \frac{1}{a b} = \frac{a + c}{a b c}$
$= \frac{2 b}{a b c}$
$= \frac{2}{a c}$

Hence proved.

So the statement is true

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Similar questions

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If a,b,c are in AP; then 1bc,1caand1ab are in AP.

The correct option is A True If a,b,c are in AP, then a +c = 2b .... (i) To check if 1bc,1caand1ab are in AP, 1bc+1ab=a+cabc =2babc =2ac Hence proved. So the statement is true

If $a, b, c$ are in AP; then $\frac{1}{b c}, \frac{1}{c a} a n d \frac{1}{a b}$ are in AP.

The correct option is A
True

If $a, b, c$ are in AP, then a +c = 2b .... (i)

To check if $\frac{1}{b c}, \frac{1}{c a} a n d \frac{1}{a b}$ are in AP,

$\frac{1}{b c} + \frac{1}{a b} = \frac{a + c}{a b c}$
$= \frac{2 b}{a b c}$
$= \frac{2}{a c}$

Hence proved.

So the statement is true