If (b + c - 2a)/a, (c + a - 2b)/b, (a + b - 2c)/c are in A.P., then show that 1/a, 1/b, 1/c are in A.P.

Question

If \(\frac{b+c−2a}{a} \), \(\frac{c+a−2b}{b} \), \(\frac{a+b−2c}{c} \) are in A.P., then show that \(\frac{1}{a}\), \(\frac{1}{b}\), \(\frac{1}{c}\) are in A.P.

Answer 1

Given, \(\frac{b+c−2a}{a}\), \(\frac{c+a−2b}{b}\), \(\frac{a+b−2c}{c}\) are in A.P

⇒ \(\bigg\{\frac{b + c − 2a}{a} + 3\bigg\}\), \(\bigg\{\frac{c + a − 2a}{b} + 3\bigg\}\), \(\bigg\{\frac{a + b − 2a}{c} + 3\bigg\}\)

are in A.P. [on adding 3 to each tech]

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

⇒ \(\frac{1}{a}\), \(\frac{1}{b}\), \(\frac{1}{c}\) are in A.P.

[Dividing each term by a + b + c]