Question

if a,b,c are in hp then prove that b+c/bc,a+c/ac,a+b/ab are in ap​

Answer 1

Given : a , b , c are in Harmonic progression (HP)

⇒ $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$

⇒ $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$  _____(1)

   

Required to Prove : (b+c) /bc ; (a+c)/ac ;(a+b)/ab   are in Arithmetic progression (AP).

$\frac{b+c}{bc} = \frac{1}{c} + \frac{1}{b} ;$  ____(i)

$\frac{a+c}{ac} = \frac{1}{c} + \frac{1}{a} ;$ _____(ii)

$\frac{a+b}{ab} = \frac{1}{b} + \frac{1}{a} ;$ _____(iii)

Checking the common difference between the successive terms,

(ii) - (i) = $[\frac{1}{c} + \frac{1}{a} ] - [ \frac{1}{c} +\frac{1}{b} ]$

⇒ $\frac{1}{a} - \frac{1}{b}$

(iii) - (ii) = $[\frac{1}{b} + \frac{1}{a} ] - [\frac{1}{c} + \frac{1}{a} ]$

⇒ $\frac{1}{b} -\frac{1}{c}$  

we know that ,

$\frac{1}{a} - \frac{1}{b} = \frac{1}{b} - \frac{1}{c}$     [∵ From (i) ]

( The difference between successive terms is equal ; there is a common difference between the terms).

⇒ (iii) - (ii) = (ii) - (i)

⇒ 2 x (ii) = (i) + (iii)

⇒ 2 x (a+c)/ac = (b+c)/bc + (a+b)/ab

∴ (b+c) /bc ; (a+c)/ac ;(a+b)/ab   are in Arithmetic progression (AP).

HOPE THIS HELPS YOU .!!

Answer 2

Step-by-step explanation:

Given : a , b , c are in Harmonic progression (HP)

⇒ \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}

b

1

−

a

1

=

c

1

−

b

1

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

b

2

=

a

1

+

c

1

_____(1)

Required to Prove : (b+c) /bc ; (a+c)/ac ;(a+b)/ab are in Arithmetic progression (AP).

\frac{b+c}{bc} = \frac{1}{c} + \frac{1}{b} ;

bc

b+c

=

c

1

+

b

1

; ____(i)

\frac{a+c}{ac} = \frac{1}{c} + \frac{1}{a} ;

ac

a+c

=

c

1

+

a

1

; _____(ii)

\frac{a+b}{ab} = \frac{1}{b} + \frac{1}{a} ;

ab

a+b

=

b

1

+

a

1

; _____(iii)

Checking the common difference between the successive terms,

(ii) - (i) = [\frac{1}{c} + \frac{1}{a} ] - [ \frac{1}{c} +\frac{1}{b} ][

c

1

+

a

1

]−[

c

1

+

b

1

]

⇒ \frac{1}{a} - \frac{1}{b}

a

1

−

b

1

(iii) - (ii) = [\frac{1}{b} + \frac{1}{a} ] - [\frac{1}{c} + \frac{1}{a} ][

b

1

+

a

1

]−[

c

1

+

a

1

]

⇒ \frac{1}{b} -\frac{1}{c}

b

1

−

c

1

we know that ,

\frac{1}{a} - \frac{1}{b} = \frac{1}{b} - \frac{1}{c}

a

1

−

b

1

=

b

1

−

c

1

[∵ From (i) ]

( The difference between successive terms is equal ; there is a common difference between the terms).

⇒ (iii) - (ii) = (ii) - (i)

⇒ 2 x (ii) = (i) + (iii)

⇒ 2 x (a+c)/ac = (b+c)/bc + (a+b)/ab

∴ (b+c) /bc ; (a+c)/ac ;(a+b)/ab are in Arithmetic progression (AP).

HOPE THIS HELPS YOU .!!