If a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in A.P, prove that a, b, c are in A.P

Solution:

It is given that a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in A.P.

Therefore,

b(1/c + 1/a) - a(1/b + 1/c) = c(1/a + 1/b) - b(1/c + 1/a)

⇒ b (a + c)/ac - a (b + c)/bc = c (a + b)/ab - b (a + c)/ac

⇒ (b²a + b²c - a²b - a²c)/abc = (c²a + c²b - b²a - b²c)/abc

⇒ b²a - a²b + b²c - a²c = c²a - b²a + c²b - b²c

⇒ ab (b - a) + c (b² - a² ) = a (c² - b² ) + bc (c - b)

⇒ ab (b - a) + c (b + a)(b - a) = a (c - b)(c + b) + bc (c - b)

⇒ (b - a)(ab + cb + ca) = (c - b)(ac + ab + bc)

⇒ b - a = c - b

Thus, a, b, c are in A.P. proved

NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 16

If a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in A.P, prove that a, b, c are in A.P

Summary:

We know that a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in A.P, using this we proved that a, b and c are in A.P