if a, b c are in hp then prove that ab/(b+c), ca/(c+a) , ab/(a+b) are in hp
Answers
Answer:
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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 .!!