If \(\frac{b+c-a}{a}\), \(\frac{c+a-b}{b}\), \(\frac{a+b-c}{c}\) a

1.	If \(\frac{b+c-a}{a}\), \(\frac{c+a-b}{b}\), \(\frac{a+b-c}{c}\) are in A.P., prove that \(\frac{1}{a}\), \(\frac{1}{b}\), \(\frac{1}{c}\) are also in A.P.
Answer» \(\frac{b+c-a}{a}\), \(\frac{c+a-b}{b}\), \(\frac{a+b-c}{c}\) are in A.P. ⇒ \(\frac{b+c-a}{a}\)+ 2, \(\frac{c+a-b}{b}\)+ 2, \(\frac{a+b-c}{c}\)+ 2 are in A.P. (Adding 2 to each term of A.P.) ⇒ \(\frac{b+c+a}{a}\), \(\frac{c+a+b}{b}\), \(\frac{a+b+c}{c}\) are in A.P. ⇒ \(\frac{1}{a}\), \(\frac{1}{b}\),\(\frac{1}{c}\) are in A.P. (Dividing each term by a + b + c)

If \(\frac{b+c-a}{a}\), \(\frac{c+a-b}{b}\), \(\frac{a+b-c}{c}\) are in A.P., prove that \(\frac{1}{a}\), \(\frac{1}{b}\), \(\frac{1}{c}\) are also in A.P.

Answer»

\(\frac{b+c-a}{a}\), \(\frac{c+a-b}{b}\), \(\frac{a+b-c}{c}\) are in A.P.

⇒ \(\frac{b+c-a}{a}\)+ 2, \(\frac{c+a-b}{b}\)+ 2, \(\frac{a+b-c}{c}\)+ 2 are in A.P. (Adding 2 to each term of A.P.)

⇒ \(\frac{b+c+a}{a}\), \(\frac{c+a+b}{b}\), \(\frac{a+b+c}{c}\) are in A.P.

⇒ \(\frac{1}{a}\), \(\frac{1}{b}\),\(\frac{1}{c}\) are in A.P. (Dividing each term by a + b + c)