Question

b+c-a/a c+a-b/b a+b-c are in ap. Proove that 1/a 1/b 1/c are in ap

Answer 1

Please see the attachment

Answer 2

Answer:

$\displaystyle \dfrac{1}{a},\dfrac{1}{b}, \dfrac{1}{c} \ are \ in \ A.P$

Step-by-step explanation:

Given,

$\displaystyle \dfrac{b+c-a}{a},\dfrac{c+a-b}{b}, \dfrac{a+b-c}{c} \ are \ in \ A.P$

Adding 2 in each term of AP:

$\displaystyle \dfrac{b+c-a}{a}+2,\dfrac{c+a-b}{b}+2, \dfrac{a+b-c}{c} +2\ are \ in \ A.P$

$\displaystyle \dfrac{b+c-a+2a}{a},\dfrac{c+a-b+2b}{b}, \dfrac{a+b-c+2c}{c} \ are \ in \ A.P$

$\displaystyle \dfrac{(a+b+c)}{a},\dfrac{(a+b+c)}{b}, \dfrac{(a+b+c)}{c} \ are \ in \ A.P$

Dividing each term by (a+b+c):

$\displaystyle \dfrac{1}{a},\dfrac{1}{b}, \dfrac{1}{c} \ are \ in \ A.P$

Hence Proved!