Question

if $a( \frac{1}{b} + \frac{1}{c} ) \: ,b( \frac{1}{c} + \frac{1}{a} ) \:, c( \frac{1}{a} + \frac{1}{b} )$ are in AP., prove that a, b and c are in AP.

This question is given in RD SHARMA book.

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Answer 1

Answer:

the answer is

Step-by-step explanation:

tex=ab/frac[1]=a=frac[2]

b=frac[1]

c=1

Answer 2

We have,

$\sf = \: a \left( \dfrac{1}{b} + \dfrac{1}{c} \right) \: ,b\left( \dfrac{1}{c} + \dfrac{1}{a}\right ) \:, c\left( \dfrac{1}{a} + \dfrac{1}{b} \right) \: are \: in \: AP$

$\sf = \: a \left( \dfrac{c + b}{bc}\right) \: ,b\left( \dfrac{a + c}{ac} \right ) \:, c\left( \dfrac{b + a}{ab} \right) \: are \: in \: AP$

$\sf = \left( \dfrac{ac + ab}{bc} \right)\: , \left(\dfrac{ab + bc}{ac}\right) , \left( \dfrac{bc + ac}{ab}\right) \: are \: in \: AP$

$\sf = \left( \dfrac{ac + ab}{bc} + 1 \right), \left(\dfrac{ab + bc}{ac} + 1\right) , \left(\dfrac{bc + ac}{ab} + 1 \right)\: are \: in \: AP$

$\sf = \left( \dfrac{ac + ab + bc}{bc} \right), \left(\dfrac{ab + bc + ac}{ac} \right) , \left(\dfrac{bc + ac + ab}{ab} \right)\: are \: in \: AP$

$\sf = \left( \dfrac{ ab + bc + ac}{bc} \right), \left(\dfrac{ab + bc + ac}{ac} \right) , \left(\dfrac{ ab + bc + ac}{ab} \right)\: are \: in \: AP$

$\sf = \left( \dfrac{ ab + bc + ac}{(ab + bc + ac) \times bc} \right), \left(\dfrac{ab + bc + ac}{(ab + bc + ac) \times ac} \right) , \left(\dfrac{ ab + bc + ac}{(ab + bc + ac) \times ab} \right)\: are \: in \: AP$

$\sf = \left( \dfrac{1}{bc} \right), \left(\dfrac{1}{ac} \right) , \left(\dfrac{1}{ab} \right)\: are \: in \: AP$

$\sf = \left( \dfrac{abc}{bc} \right), \left(\dfrac{abc}{ac} \right) , \left(\dfrac{abc}{ab} \right)\: are \: in \: AP$

$\sf = \left( a \right), \left(b \right) , \left(c \right)\: are \: in \: AP$

Hence proved

Concept :-

The only concept used in this question is that we are allowed to do addition, subtraction, multiplication and division with each term of an AP. AP will remain AP if we do any of the operator mentioned above.